description: 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].
solution: #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;
  }
  