#include <bits/stdc++.h>
using namespace std;
const long long dv = 1e9 + 7;
long long n, m, d, ps[200069], fq[200069], pr[200069][18], dp[200069][18];
int main() {
  long long i, j, p, sm, z;
  scanf("%lld%lld%lld", &n, &m, &d);
  for (i = 1; i <= n; i++) {
    scanf("%lld", ps + i);
    ps[i] += ps[i - 1];
  }
  z = ps[n] * m % d == 0;
  fq[0]++;
  for (i = 1; i < n; i++) {
    z = (z + (fq[ps[i] % d] + fq[(ps[i] + d - ps[n] * m % d) % d]) * m) % dv;
    fq[ps[i] % d]++;
  }
  for (j = 0; j < d; j++) {
    pr[j][0] = (j + ps[n]) % d;
    dp[j][0] = fq[j];
  }
  for (i = 1; i < 18; i++) {
    for (j = 0; j < d; j++) {
      pr[j][i] = pr[pr[j][i - 1]][i - 1];
      dp[j][i] = (dp[j][i - 1] + dp[pr[j][i - 1]][i - 1]) % dv;
    }
  }
  for (i = 0; i < d; i++) {
    p = i;
    sm = 0;
    for (j = 0; j < 18; j++) {
      if (m - 1 >> j & 1) {
        sm = (sm + dp[p][j]) % dv;
        p = pr[p][j];
      }
    }
    z = (z + sm * fq[p] % dv * m) % dv;
  }
  printf("%lld\n", z);
}

### Prompt

Create a solution in Cpp for the following problem:

Denote a cyclic sequence of size n as an array s such that s_n is adjacent to s_1. The segment s[r, l] where l < r is the concatenation of s[r, n] and s[1, l].

You are given an array a consisting of n integers. Define b as the cyclic sequence obtained from concatenating m copies of a. Note that b has size n ⋅ m.

You are given an integer k where k = 1 or k is a prime number. Find the number of different segments in b where the sum of elements in the segment is divisible by k.

Two segments are considered different if the set of indices of the segments are different. For example, when n = 3 and m = 2, the set of indices for segment s[2, 5] is \{2, 3, 4, 5\}, and for segment s[5, 2] is \{5, 6, 1, 2\}. In particular, the segments s[1, 6], s[2,1], …, s[6, 5] are considered as the same segment.

Output the answer modulo 10^9 + 7.

Input

The first line contains three integers n, m, and k (1 ≤ n, m, k ≤ 2 ⋅ 10^5, k = 1 or k is a prime number).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 2 ⋅ 10^5).

Output

Output an integer denoting the number of different segments in b where the sum of elements in the segment is divisible by k, modulo 10^9 + 7.

Examples

Input


5 1 5
1 2 3 4 3


Output


4


Input


5 1 5
1 2 3 4 5


Output


5


Input


5 4 5
1 2 3 4 5


Output


125

Note

In the first example, all valid segments are [1,4], [2, 3], [3, 5], and [4, 2].

In the second example, one of the valid segments is [1, 5].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long dv = 1e9 + 7;
long long n, m, d, ps[200069], fq[200069], pr[200069][18], dp[200069][18];
int main() {
  long long i, j, p, sm, z;
  scanf("%lld%lld%lld", &n, &m, &d);
  for (i = 1; i <= n; i++) {
    scanf("%lld", ps + i);
    ps[i] += ps[i - 1];
  }
  z = ps[n] * m % d == 0;
  fq[0]++;
  for (i = 1; i < n; i++) {
    z = (z + (fq[ps[i] % d] + fq[(ps[i] + d - ps[n] * m % d) % d]) * m) % dv;
    fq[ps[i] % d]++;
  }
  for (j = 0; j < d; j++) {
    pr[j][0] = (j + ps[n]) % d;
    dp[j][0] = fq[j];
  }
  for (i = 1; i < 18; i++) {
    for (j = 0; j < d; j++) {
      pr[j][i] = pr[pr[j][i - 1]][i - 1];
      dp[j][i] = (dp[j][i - 1] + dp[pr[j][i - 1]][i - 1]) % dv;
    }
  }
  for (i = 0; i < d; i++) {
    p = i;
    sm = 0;
    for (j = 0; j < 18; j++) {
      if (m - 1 >> j & 1) {
        sm = (sm + dp[p][j]) % dv;
        p = pr[p][j];
      }
    }
    z = (z + sm * fq[p] % dv * m) % dv;
  }
  printf("%lld\n", z);
}

```