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 = int(2e5) + 5;
  const int M = 1000000007;
  int ans, n, m, k, r;
  int a[N], b[N], p[N], s[N];
  int cnt[19][N];
  void add(int &x, int y) {
    x += y;
    if (x >= M) x -= M;
  }
  void sub(int &x, int y) {
    x -= y;
    if (x < 0) x += M;
  }
  int main() {
    cin >> n >> m >> k;
    for (int i = 0; i < n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) b[i] = (b[i - 1] + a[i - 1]) % k;
    for (int i = 0; i < n; ++i) cnt[0][b[i]]++;
    r = p[0] = b[n];
    for (int i = 1; i <= 17; ++i) {
      for (int j = 0; j < k; ++j) {
        add(cnt[i][j], cnt[i - 1][j]);
        add(cnt[i][(j + r) % k], cnt[i - 1][j]);
      }
      r = r * 2 % k;
      p[i] = r;
    }
    r = 0;
    for (int i = 17; i >= 0; --i)
      if ((m >> i) & 1) {
        for (int j = 0; j < k; ++j) add(s[(j + r) % k], cnt[i][j]);
        r = (r + p[i]) % k;
      }
    ans = 0;
    for (int i = 0; i < n; ++i) {
      sub(s[b[i]], 1);
      add(ans, s[b[i]]);
      add(s[(b[i] + r) % k], 1);
    }
    ans = 1ll * ans * m % M;
    if (!r) add(ans, 1);
    cout << ans << endl;
    return 0;
  }
  