bzoj4870

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 60;
struct mat {
    ll a[N][N];
} x, g;
ll n, k, p, r;
void build()
{
    x.a[0][0] = 1; //0个物品选0个的方案数为1 
    for(int i = 0; i < k; ++i) 
    {
        ++g.a[i][i]; ++g.a[i][(i + 1) % k]; //矩阵的系数 f[i][j]=f[i-1][j]+f[i-1][j-1] 所以j=1,j-1=1 
    }
}
mat operator * (const mat &A, const mat &B)
{
    mat ret; memset(ret.a, 0, sizeof(ret.a));
    for(int i = 0; i < k; ++i)
        for(int j = 0; j < k; ++j)
            for(int x = 0; x < k; ++x) ret.a[i][j] = (ret.a[i][j] + A.a[i][x] * B.a[x][j]) % p;
    return ret;
}
void power(mat A, ll t)
{
    for(; t; A = A * A, t >>= 1) if(t & 1) x = x * A;
}
int main()
{
    scanf("%lld%lld%lld%lld", &n, &p, &k, &r);
    build();
    power(g, n * k);
    printf("%lld\n", x.a[0][r]);
    return 0;
}