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 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);
  }
  