#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, P = 1e9 + 7;
int n, m, p, f[N][20], a[N], q[N], s[20], b[N];
inline void add(int &x, int y) {
  x = (x + y) % P;
  if (x < 0) x += P;
}
int main() {
  scanf("%d%d%d", &n, &m, &p);
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
  for (int i = (1); i <= (n); i++) q[i] = (q[i - 1] + a[i]) % p;
  for (int i = (0); i <= (n - 1); i++) f[q[i]][0]++;
  int d = q[n], ans = 0;
  s[0] = d;
  for (int j = (1); j <= (17); j++) {
    for (int i = (0); i <= (p - 1); i++) {
      add(f[i][j], f[i][j - 1]);
      add(f[(i + d) % p][j], f[i][j - 1]);
    }
    d = d * 2 % p, s[j] = d;
  }
  d = 0;
  for (int j = (17); j >= (0); j--)
    if (m & (1 << j)) {
      for (int i = (0); i <= (p - 1); i++) add(b[(i + d) % p], f[i][j]);
      d = (d + s[j]) % p;
    }
  for (int i = (0); i <= (n - 1); i++) {
    add(b[q[i]], -1), add(ans, b[q[i]]);
    add(b[(d + q[i]) % p], 1);
  }
  ans = 1ll * ans * m % P;
  printf("%d", (ans + (d == 0)) % P);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 int N = 2e5 + 5, P = 1e9 + 7;
int n, m, p, f[N][20], a[N], q[N], s[20], b[N];
inline void add(int &x, int y) {
  x = (x + y) % P;
  if (x < 0) x += P;
}
int main() {
  scanf("%d%d%d", &n, &m, &p);
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
  for (int i = (1); i <= (n); i++) q[i] = (q[i - 1] + a[i]) % p;
  for (int i = (0); i <= (n - 1); i++) f[q[i]][0]++;
  int d = q[n], ans = 0;
  s[0] = d;
  for (int j = (1); j <= (17); j++) {
    for (int i = (0); i <= (p - 1); i++) {
      add(f[i][j], f[i][j - 1]);
      add(f[(i + d) % p][j], f[i][j - 1]);
    }
    d = d * 2 % p, s[j] = d;
  }
  d = 0;
  for (int j = (17); j >= (0); j--)
    if (m & (1 << j)) {
      for (int i = (0); i <= (p - 1); i++) add(b[(i + d) % p], f[i][j]);
      d = (d + s[j]) % p;
    }
  for (int i = (0); i <= (n - 1); i++) {
    add(b[q[i]], -1), add(ans, b[q[i]]);
    add(b[(d + q[i]) % p], 1);
  }
  ans = 1ll * ans * m % P;
  printf("%d", (ans + (d == 0)) % P);
  return 0;
}

```