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 MAXN = 200005;
  const double PI = acos(-1);
  const long long MOD = 1000000007;
  struct Complex {
    double real, imag;
    Complex(double R = 0, double I = 0) {
      real = R;
      imag = I;
    }
    Complex operator+(const Complex &p) const {
      return Complex(real + p.real, imag + p.imag);
    }
    Complex operator-(const Complex &p) const {
      return Complex(real - p.real, imag - p.imag);
    }
    Complex operator*(const Complex &p) const {
      return Complex(real * p.real - imag * p.imag,
                     real * p.imag + imag * p.real);
    }
  } a[MAXN << 2], b[MAXN << 2], wn[2][MAXN << 2];
  int n, m, k, N;
  int c[MAXN];
  int d[MAXN];
  int cnt2[MAXN];
  int cnt3[MAXN];
  int rev[MAXN << 2];
  long long ans;
  long long cnt[MAXN];
  void FFTinit(int l) {
    N = 1;
    while (N < l) N <<= 1;
    for (int i = 1; i < N; i++) rev[i] = (rev[i >> 1] >> 1) + (i & 1) * (N >> 1);
    int half = N >> 1;
    for (int i = 0; i < half; i++) {
      wn[0][half + i] = Complex(cos(PI * i / half), sin(PI * i / half));
      wn[1][half + i] = Complex(cos(PI * i / half), -sin(PI * i / half));
    }
    for (int i = half - 1; i >= 0; i--) {
      wn[0][i] = wn[0][i << 1];
      wn[1][i] = wn[1][i << 1];
    }
  }
  void FFT(Complex *v, int inv) {
    for (int i = 0; i < N; i++)
      if (i < rev[i]) swap(v[i], v[rev[i]]);
    for (int i = 1; i < N; i <<= 1)
      for (int j = 0; j < N; j += (i << 1)) {
        Complex *l = v + j, *r = v + j + i, *w = wn[inv == -1] + i, A;
        for (int k = 0; k < i; k++, l++, r++, w++) {
          A = *w * *r;
          *r = *l - A;
          *l = *l + A;
        }
      }
    if (inv == -1) {
      Complex Inv(1.0 / N, 0);
      for (int i = 0; i < N; i++) v[i] = v[i] * Inv;
    }
  }
  int main() {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &c[i]);
      d[i] = (d[i - 1] + c[i]) % k;
    }
    int tot = (long long)d[n] * m % k;
    for (int i = 1; i <= n; i++) a[d[i]].real++;
    for (int i = 1; i < m; i++) b[(long long)i * d[n] % k].real++;
    FFTinit(k << 1 | 1);
    FFT(a, 1);
    FFT(b, 1);
    for (int i = 0; i < N; i++) a[i] = a[i] * b[i];
    FFT(a, -1);
    for (int i = 0; i <= k * 2; i++) cnt[i % k] += (long long)(a[i].real + 0.5);
    for (int i = 0; i < k; i++) cnt[i] %= MOD;
    for (int i = 1; i <= n; i++) cnt2[d[i]]++;
    for (int i = 1; i <= n; i++) {
      (ans += cnt2[d[i - 1]] + cnt[d[i - 1]] + cnt3[(d[i - 1] - tot + k) % k]) %=
          MOD;
      cnt2[d[i]]--;
      if (i > 1) cnt3[d[i - 1]]++;
    }
    (ans *= m) %= MOD;
    if (!tot) (ans -= m - 1) %= MOD;
    printf("%lld\n", ans);
    return 0;
  }
  