codeforces621E————Wet Shark and Blocks(动态规划，矩阵快速幂)

E. Wet Shark and Blocks

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are b blocks of digits. Each one consisting of the same n digits, which are given to you in the input. Wet Shark must choose exactly one digit from each block and concatenate all of those digits together to form one large integer. For example, if he chooses digit 1 from the first block and digit 2 from the second block, he gets the integer 12. 

Wet Shark then takes this number modulo x. Please, tell him how many ways he can choose one digit from each block so that he gets exactly k as the final result. As this number may be too large, print it modulo 109 + 7.

Note, that the number of ways to choose some digit in the block is equal to the number of it's occurrences. For example, there are 3 ways to choose digit 5 from block 3 5 6 7 8 9 5 1 1 1 1 5.

Input

The first line of the input contains four space-separated integers, n, b, k and x (2 ≤ n ≤ 50 000, 1 ≤ b ≤ 109, 0 ≤ k < x ≤ 100, x ≥ 2) — the number of digits in one block, the number of blocks, interesting remainder modulo x and modulo x itself.

The next line contains n space separated integers ai (1 ≤ ai ≤ 9), that give the digits contained in each block.

Output

Print the number of ways to pick exactly one digit from each blocks, such that the resulting integer equals k modulo x.

Examples

input

12 1 5 10
3 5 6 7 8 9 5 1 1 1 1 5

output

3

input

3 2 1 2
6 2 2

output

0

input

3 2 1 2
3 1 2

output

6

Note

In the second sample possible integers are 22, 26, 62 and 66. None of them gives the remainder 1 modulo 2.

In the third sample integers 11, 13, 21, 23, 31 and 33 have remainder 1 modulo 2. There is exactly one way to obtain each of these integers, so the total answer is 6.

有b个包括相同的n个数字的block，你需要从每个block中选出1个组成一个数，问组成的数%x=k的方案数

首先，考虑dp递推方程，dp[i][j]表示考虑i个block，模x等于j的方案数。dp[i][j*10+y]=dp[I][j]*num[y].(其中y可以是从1到9九个数，num[y]表示block中数字y的个数)

然而因为b太大，无法直接递推，所以考虑用矩阵快速幂优化递推关系式。

因为得到的数是模x，所以一共有0到x-1的x种可能性。

用一个矩阵x*1的矩阵存结果，表示这x种答案的方案数，矩阵第一个数为1，其余为0

然后构造系数矩阵，显然系数矩阵第i行第j列代表着通过一个block把答案从j变成i的方案数。所以只要先求出系数矩阵再用矩阵快速幂即可。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long LL;

const int MAXN = 1000010;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;

typedef vector<int> vec;
typedef vector<vec> mat;
mat mat_mul(mat &A, mat &B) {
	mat C(A.size(), vec(B[0].size()));
	for(int i = 0; i < A.size(); i++) {
		for(int j = 0; j < B[0].size(); j++) {
			for(int k = 0; k < B.size(); k++) {
				C[i][j] = ((LL)A[i][k] * B[k][j] + C[i][j]) % mod;
			}
		}
	}
	return C;
}
mat mat_pow(mat A, LL n) {
	mat B(A.size(), vec(A.size()));
	for(int i = 0; i < A.size(); i++) B[i][i] = 1;
	while(n) {
		if(n & 1) B = mat_mul(B, A);
		A = mat_mul(A, A);
		n >>= 1;
	}
	return B;
}

int cnt[10];
int main() {
	int n, b, k, x; //FIN;
	scanf("%d%d%d%d", &n, &b, &k, &x);
	for(int i = 1; i <= n; i++) {
		int t; scanf("%d", &t);
		cnt[t]++;
	}
	mat A(x, vec(x)), B(x, vec(1));
	B[0][0] = 1;
	for(int i = 0; i < x; i++) {
		for(int j = 1; j <= 9; j++) {
			A[(i * 10 + j) % x][i] += cnt[j];
		}
	}
	A = mat_pow(A, b);
	B = mat_mul(A, B);
	printf("%d\n", B[k][0]);
	return 0;
}