#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;
}

### Prompt

Generate a Cpp solution to the following problem:

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].

### Response

```cpp
#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;
}

```