In this problem, the first confusing thing is that, where is the MOD value !! But if you carefully read the problem statement, the MOD value is there indirectly as input ! 😀
When we have to find the last 1 digit of a given integer n, then the last digit k is-
k=n%10
or, k=n%10^1
Similarly when we are to find the last 1 digit of a given integer n, then –
k=n%100
or, k=n%10^2
similarly for finding the last m digit of a given integer n, then –
or, k=n%10^m
And here, this 10^m is the MOD value !! 😀

#include <bits/stdc++.h>

#define pf                  printf
#define sf(a)               scanf("%d",&a)
#define sfl(a)              scanf("%lld",&a)
#define sff(a,b)            scanf("%d %d",&a,&b)
#define sffl(a,b)           scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)         scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)        scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d)      scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d)     scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e)   scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e)  scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a)              scanf("%c",&a)
#define ms(a,b)             memset(a,b,sizeof(a))
#define pb(a)               push_back(a)
#define pbp(a,b)            push_back({a,b})
#define db                  double
#define ft                  float
#define ll                  long long
#define ull                 unsigned long long
#define ff                  first
#define ss                  second
#define sz(x)               x.size()
#define qu                  queue
#define pqu                 priority_queue
#define vc                  vector
#define vi                  vector<int>
#define vll                 vector<long long>
#define pii                 pair<int,int>
#define pis                 pair<int,string>
#define psi                 pair<string,int>
#define all(x)              x.begin(),x.end()
#define CIN                 ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c)       max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c)       min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define loop0(i,n)          for(int i=0;i<n;i++)
#define loop1(i,n)          for(int i=1;i<=n;i++)
#define loopcmp(i,n)        for(int i=1;i<n;i++)
#define looprev(i,n)        for(int i=n-1; i>=0; i--)
#define loopab(i,a,b)       for(int i=a;i<=b;i++)
#define loopba(i,b,a)       for(int i=b;i>=a;i--)
#define stlloop(x)          for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b)           __gcd(a, b)
#define lcm(a, b)           ((a)*((b)/gcd(a,b)))
#define case1(z)            cout<<"Case "<<z<<": "
#define case2(z)            printf("Case %d: ",z)
#define PI                  acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty)        tx>=0 && tx<row && ty>=0 && ty<col
#define intlim              2147483648
#define MAX                 1000000
#define inf                 100000000

/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*---------------------------------------------------------------------*/

using namespace std;

/*----------------------Matrix-----------------------*/

ll MOD;

struct matrix
{
    ll mat[2][2];
    int row,col;

    matrix()
    {
        ms(mat,0);
    }
    matrix(int a, int b)
    {
        row=a,col=b;
        ms(mat,0);
    }

    matrix operator*(const matrix &p) const
    {
        assert(col == p.row);
        matrix result;
        result.row = row;
        result.col = p.col;
        for (int i = 0; i < result.row; i++)
        {
            for (int j = 0; j < result.col; j++)
            {
                ll sum = 0;
                for (int k = 0; k <col;  k++)
                {
                    sum += ((mat[i][k]%MOD) * (p.mat[k][j]%MOD))%MOD;
                    sum%=MOD;
                }
                result.mat[i][j] = sum;
            }
        }
        return result;
    }
    matrix operator+ (const matrix &p) const
    {
        assert(row==p.row && col==p.col);
        matrix result;
        result.row=row;
        result.col=col;
        for(int i=0; i<result.row; i++)
        {
            for(int j=0; j<result.col; j++)
                result.mat[i][j]=((mat[i][j]%MOD)+(p.mat[i][j]%MOD))%MOD;;
        }
        return result;
    }

    matrix identity()
    {
        matrix result;
        result.row=row;
        result.col=col;
        for(int i=0; i<row; i++)
            result.mat[i][i]=1;
        return result;
    }

    matrix pow(ll pow)
    {
        matrix result=(*this);
        matrix ret=(*this).identity();
        while(pow)
        {
            if(pow % 2==1)
                ret=ret*result;
            result=result*result;
            pow/=2;
        }
        return ret;
    }

    void show()
    {
        printf("-----------------------------\n");
        for(int i=0; i<row; i++)
        {
            for(int j=0; j<col; j++)
                printf("%lld ",mat[i][j]);
            printf("\n");
        }
        printf("-----------------------------\n");
    }

};

ll power(int n,int p)
{
    if(p==0) return 1;
    ll r=1;
    for(int i=0;i<p;i++) r*=n;
    return r;
}

/*--------------------------Matrix End---------------------*/


int main()
{

    ///freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    ll a,b,n,m,t;
    sfl(t);
    loop1(z,t)
    {
        sffffl(a,b,n,m);
        MOD=power(10,m);

        matrix base(2,2), multiplicand(2,1);

        base.mat[0][0]=base.mat[0][1]=base.mat[1][0]=1;
        multiplicand.mat[0][0]=b;
        multiplicand.mat[1][0]=a;

        base=base.pow(n-1);

        base=base*multiplicand;

        ll rs=base.mat[0][0];

        if(n==0) rs=a%MOD;
        else if(n==1) rs=b%MOD;
        case1(z);
        cout<<rs<<endl;
    }
    return 0;
}