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 double PI = acos(-1);
  int base = 1;
  vector<complex<double>> roots = {{0, 0}, {1, 0}};
  vector<int> rev = {0, 1};
  vector<complex<double>> f;
  void ensure_base(int nbase) {
    if (nbase <= base) {
      return;
    }
    rev.resize(1 << nbase);
    for (int i = 0; i < (1 << nbase); i++) {
      rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (nbase - 1));
    }
    roots.resize(1 << nbase);
    while (base < nbase) {
      double angle = 2 * PI / (1 << (base + 1));
      for (int i = 1 << (base - 1); i < (1 << base); i++) {
        roots[i << 1] = roots[i];
        double angle_i = angle * (2 * i + 1 - (1 << base));
        roots[(i << 1) + 1] = complex<double>(cos(angle_i), sin(angle_i));
      }
      base++;
    }
  }
  void fft(vector<complex<double>>& a, int n) {
    int zeros = __builtin_ctz(n);
    ensure_base(zeros);
    int shift = base - zeros;
    for (int i = 0; i < n; i++) {
      if (i < (rev[i] >> shift)) {
        swap(a[i], a[rev[i] >> shift]);
      }
    }
    for (int k = 1; k < n; k <<= 1) {
      for (int i = 0; i < n; i += 2 * k) {
        for (int j = 0; j < k; j++) {
          complex<double> z = a[i + j + k] * roots[j + k];
          a[i + j + k] = a[i + j] - z;
          a[i + j] = a[i + j] + z;
        }
      }
    }
  }
  vector<long long> multiply(const vector<int>& a, const vector<int>& b) {
    int need = (int)a.size() + (int)b.size() - 1;
    int nbase = 1;
    while ((1 << nbase) < need) {
      nbase++;
    }
    ensure_base(nbase);
    int sz = 1 << nbase;
    if (sz > (int)f.size()) {
      f.resize(sz);
    }
    for (int i = 0; i < sz; i++) {
      int x = (i < (int)a.size() ? a[i] : 0);
      int y = (i < (int)b.size() ? b[i] : 0);
      f[i] = complex<double>(x, y);
    }
    fft(f, sz);
    complex<double> r(0, -0.25 / (sz >> 1));
    for (int i = 0; i <= (sz >> 1); i++) {
      int j = (sz - i) & (sz - 1);
      complex<double> z = (f[j] * f[j] - conj(f[i] * f[i])) * r;
      f[j] = (f[i] * f[i] - conj(f[j] * f[j])) * r;
      f[i] = z;
    }
    for (int i = 0; i < (sz >> 1); i++) {
      complex<double> a0 = (f[i] + f[i + (sz >> 1)]) * complex<double>(0.5, 0);
      complex<double> a1 = (f[i] - f[i + (sz >> 1)]) * complex<double>(0.5, 0) *
                           roots[(sz >> 1) + i];
      f[i] = a0 + a1 * complex<double>(0, 1);
    }
    fft(f, sz >> 1);
    vector<long long> res(need);
    for (int i = 0; i < need; i++) {
      res[i] = llround(i % 2 == 0 ? f[i >> 1].real() : f[i >> 1].imag());
    }
    return res;
  }
  template <long long mod>
  struct modular {
    long long value;
    modular(long long x = 0) {
      value = x % mod;
      if (value < 0) value += mod;
    }
    modular& operator+=(const modular& other) {
      if ((value += other.value) >= mod) value -= mod;
      return *this;
    }
    modular& operator-=(const modular& other) {
      if ((value -= other.value) < 0) value += mod;
      return *this;
    }
    modular& operator*=(const modular& other) {
      value = value * other.value % mod;
      return *this;
    }
    modular& operator/=(const modular& other) {
      long long a = 0, b = 1, c = other.value, m = mod;
      while (c != 0) {
        long long t = m / c;
        m -= t * c;
        swap(c, m);
        a -= t * b;
        swap(a, b);
      }
      a %= mod;
      if (a < 0) a += mod;
      value = value * a % mod;
      return *this;
    }
    modular operator+(const modular& rhs) const { return modular(*this) += rhs; }
    modular operator-(const modular& rhs) const { return modular(*this) -= rhs; }
    modular operator*(const modular& rhs) const { return modular(*this) *= rhs; }
    modular operator/(const modular& rhs) const { return modular(*this) /= rhs; }
    modular& operator++() { return *this += 1; }
    modular& operator--() { return *this -= 1; }
    modular operator++(int) {
      modular res(*this);
      *this += 1;
      return res;
    }
    modular operator--(int) {
      modular res(*this);
      *this -= 1;
      return res;
    }
    modular operator-() const { return modular(-value); }
    bool operator==(const modular& rhs) const { return value == rhs.value; }
    bool operator!=(const modular& rhs) const { return value != rhs.value; }
    bool operator<(const modular& rhs) const { return value < rhs.value; }
  };
  template <long long mod>
  string to_string(const modular<mod>& x) {
    return to_string(x.value);
  }
  template <long long mod>
  ostream& operator<<(ostream& stream, const modular<mod>& x) {
    return stream << x.value;
  }
  template <long long mod>
  istream& operator>>(istream& stream, modular<mod>& x) {
    stream >> x.value;
    x.value %= mod;
    if (x.value < 0) x.value += mod;
    return stream;
  }
  constexpr long long mod = (long long)1e9 + 7;
  using mint = modular<mod>;
  mint power(mint a, long long n) {
    mint res = 1;
    while (n > 0) {
      if (n & 1) {
        res *= a;
      }
      a *= a;
      n >>= 1;
    }
    return res;
  }
  vector<mint> fact(1, 1);
  vector<mint> finv(1, 1);
  mint C(int n, int k) {
    if (n < k || k < 0) {
      return mint(0);
    }
    while ((int)fact.size() < n + 1) {
      fact.emplace_back(fact.back() * (int)fact.size());
      finv.emplace_back(mint(1) / fact.back());
    }
    return fact[n] * finv[k] * finv[n - k];
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m, k;
    cin >> n >> m >> k;
    vector<int> a(n);
    int b = 0;
    vector<int> c(k);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      a[i] %= k;
      b = (b + a[i]) % k;
      c[b]++;
    }
    vector<int> d(k);
    int e = 0;
    for (int i = 0; i < m; i++) {
      d[e]++;
      e = (e + b) % k;
    }
    auto z = multiply(c, d);
    for (int i = 0; i < k - 1; i++) {
      z[i] += z[i + k];
    }
    z.resize(k);
    ;
    mint ans = 0;
    int g = (int)((1LL * b * m) % k);
    b = 0;
    for (int i = 0; i < n; i++) {
      if (i) {
        z[b]--;
        z[(b + g) % k]++;
      }
      ans += z[b];
      b = (b + a[i]) % k;
    }
    if (g == 0) {
      ans -= n;
    }
    ans *= m;
    if (g == 0) {
      ans++;
    }
    cout << ans << '\n';
    return 0;
  }
  