#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
int ad(int k1, int k2) { return k1 += k2 - P, k1 += k1 >> 31 & P; }
int su(int k1, int k2) { return k1 -= k2, k1 += k1 >> 31 & P; }
int mu(int k1, int k2) { return 1LL * k1 * k2 % P; }
void uad(int& k1, int k2) { k1 += k2 - P, k1 += k1 >> 31 & P; }
void usu(int& k1, int k2) { k1 -= k2, k1 += k1 >> 31 & P; }
template <class... T>
int ad(int k1, T... k2) {
  return ad(k1, ad(k2...));
}
template <class... T>
int su(int k1, T... k2) {
  return su(k1, ad(k2...));
}
template <class... T>
int mu(int k1, T... k2) {
  return mu(k1, mu(k2...));
}
template <class... T>
void uad(int& k1, T... k2) {
  return uad(k1, ad(k2...));
}
template <class... T>
void usu(int& k1, T... k2) {
  return usu(k1, ad(k2...));
}
int po(int k1, int k2) {
  int k3 = 1;
  for (; k2; k2 >>= 1, k1 = mu(k1, k1))
    if (k2 & 1) k3 = mu(k3, k1);
  return k3;
}
const int N = 100005;
int n, K, f[N], a[N], c[N], tot, fac[N], inv[N], ifac[N];
map<int, int> b;
int C(int k1, int k2) {
  if (k1 < 0 || k2 < 0 || k1 - k2 < 0) return 0;
  return 1LL * fac[k1] * ifac[k2] % P * ifac[k1 - k2] % P;
}
int main() {
  fac[0] = fac[1] = inv[0] = inv[1] = ifac[0] = ifac[1] = 1;
  for (int i = 2; i < N; ++i)
    fac[i] = 1LL * fac[i - 1] * i % P,
    inv[i] = 1LL * (P - P / i) * inv[P % i] % P,
    ifac[i] = 1LL * ifac[i - 1] * inv[i] % P;
  scanf("%d%d", &n, &K);
  f[0] = 0;
  f[1] = 1;
  for (int x = (1); x <= (n - 1); ++x) {
    int rhs = (n + 1LL * n * K % P * f[x] - 1LL * x * (K - 1) % P * f[x - 1] -
               (x + 1LL * (n - x) * (K - 1)) % P * f[x]) %
              P;
    if (rhs < P) rhs += P;
    f[x + 1] = mu(rhs, po(n - x, P - 2));
  }
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i]);
    if (a[i] == -1)
      ++tot;
    else
      ++b[a[i]];
  }
  int ans = 0;
  for (auto x : b) {
    ++c[x.second];
  }
  c[0] += K - ((int)(b).size());
  const int invK = po(K, P - 2);
  for (int i = (0); i <= (n); ++i)
    if (c[i]) {
      for (int j = (0); j <= (tot); ++j) {
        uad(ans, mu(c[i], f[i + j], C(tot, j), po(invK, j),
                    po(su(1, invK), tot - j)));
      }
    }
  printf("%d\n", su(f[n], ans));
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

Mr. Chanek opened a letter from his fellow, who is currently studying at Singanesia. Here is what it says.

Define an array b (0 ≤ b_i < k) with n integers. While there exists a pair (i, j) such that b_i ≠ b_j, do the following operation:

  * Randomly pick a number i satisfying 0 ≤ i < n. Note that each number i has a probability of 1/n to be picked. 
  * Randomly Pick a number j satisfying 0 ≤ j < k. 
  * Change the value of b_i to j. It is possible for b_i to be changed to the same value. 



Denote f(b) as the expected number of operations done to b until all elements of b are equal. 

You are given two integers n and k, and an array a (-1 ≤ a_i < k) of n integers. 

For every index i with a_i = -1, replace a_i with a random number j satisfying 0 ≤ j < k. Let c be the number of occurrences of -1 in a. There are k^c possibilites of a after the replacement, each with equal probability of being the final array.

Find the expected value of f(a) modulo 10^9 + 7. 

Formally, let M = 10^9 + 7. It can be shown that the answer can be expressed as an irreducible fraction p/q, where p and q are integers and q not ≡ 0 \pmod{M}. Output the integer equal to p ⋅ q^{-1} mod M. In other words, output such an integer x that 0 ≤ x < M and x ⋅ q ≡ p \pmod{M}.

After reading the letter, Mr. Chanek gave the task to you. Solve it for the sake of their friendship!

Input

The first line contains two integers n and k (2 ≤ n ≤ 10^5, 2 ≤ k ≤ 10^9). 

The second line contains n integers a_1, a_2, …, a_n (-1 ≤ a_i < k).

Output

Output an integer denoting the expected value of f(a) modulo 10^9 + 7.

Examples

Input


2 2
0 1


Output


2


Input


2 2
0 -1


Output


1


Input


3 3
0 1 1


Output


12


Input


3 3
-1 -1 -1


Output


11


Input


10 9
-1 0 -1 1 1 2 2 3 3 3


Output


652419213

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
int ad(int k1, int k2) { return k1 += k2 - P, k1 += k1 >> 31 & P; }
int su(int k1, int k2) { return k1 -= k2, k1 += k1 >> 31 & P; }
int mu(int k1, int k2) { return 1LL * k1 * k2 % P; }
void uad(int& k1, int k2) { k1 += k2 - P, k1 += k1 >> 31 & P; }
void usu(int& k1, int k2) { k1 -= k2, k1 += k1 >> 31 & P; }
template <class... T>
int ad(int k1, T... k2) {
  return ad(k1, ad(k2...));
}
template <class... T>
int su(int k1, T... k2) {
  return su(k1, ad(k2...));
}
template <class... T>
int mu(int k1, T... k2) {
  return mu(k1, mu(k2...));
}
template <class... T>
void uad(int& k1, T... k2) {
  return uad(k1, ad(k2...));
}
template <class... T>
void usu(int& k1, T... k2) {
  return usu(k1, ad(k2...));
}
int po(int k1, int k2) {
  int k3 = 1;
  for (; k2; k2 >>= 1, k1 = mu(k1, k1))
    if (k2 & 1) k3 = mu(k3, k1);
  return k3;
}
const int N = 100005;
int n, K, f[N], a[N], c[N], tot, fac[N], inv[N], ifac[N];
map<int, int> b;
int C(int k1, int k2) {
  if (k1 < 0 || k2 < 0 || k1 - k2 < 0) return 0;
  return 1LL * fac[k1] * ifac[k2] % P * ifac[k1 - k2] % P;
}
int main() {
  fac[0] = fac[1] = inv[0] = inv[1] = ifac[0] = ifac[1] = 1;
  for (int i = 2; i < N; ++i)
    fac[i] = 1LL * fac[i - 1] * i % P,
    inv[i] = 1LL * (P - P / i) * inv[P % i] % P,
    ifac[i] = 1LL * ifac[i - 1] * inv[i] % P;
  scanf("%d%d", &n, &K);
  f[0] = 0;
  f[1] = 1;
  for (int x = (1); x <= (n - 1); ++x) {
    int rhs = (n + 1LL * n * K % P * f[x] - 1LL * x * (K - 1) % P * f[x - 1] -
               (x + 1LL * (n - x) * (K - 1)) % P * f[x]) %
              P;
    if (rhs < P) rhs += P;
    f[x + 1] = mu(rhs, po(n - x, P - 2));
  }
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i]);
    if (a[i] == -1)
      ++tot;
    else
      ++b[a[i]];
  }
  int ans = 0;
  for (auto x : b) {
    ++c[x.second];
  }
  c[0] += K - ((int)(b).size());
  const int invK = po(K, P - 2);
  for (int i = (0); i <= (n); ++i)
    if (c[i]) {
      for (int j = (0); j <= (tot); ++j) {
        uad(ans, mu(c[i], f[i + j], C(tot, j), po(invK, j),
                    po(su(1, invK), tot - j)));
      }
    }
  printf("%d\n", su(f[n], ans));
  return 0;
}

```