#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010, P = 1000000007;
int n, m, K, f[maxn], p[maxn], s[maxn], fact[maxn], finv[maxn];
unordered_map<int, int> cnt;
int qp(int x, int y) {
  int z = 1;
  for (; y; y >>= 1, x = 1LL * x * x % P) {
    if (y & 1) z = 1LL * z * x % P;
  }
  return z;
}
int binom(int n, int m) {
  return 1LL * fact[n] * finv[m] % P * finv[n - m] % P;
}
int solve(int x) {
  static unordered_map<int, int> mp;
  if (mp.count(x)) return mp[x];
  int ans = 0;
  for (int i = 0; i <= m; i++) ans = (ans + 1LL * p[i] * s[i + x]) % P;
  return mp[x] = ans;
}
int main() {
  scanf("%d %d", &n, &K);
  for (int i = fact[0] = finv[0] = 1; i <= n; i++) {
    fact[i] = 1LL * i * fact[i - 1] % P;
    finv[i] = qp(fact[i], P - 2);
  }
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x), cnt[x]++;
  }
  f[0] = K - 1;
  for (int i = 1; i < n; i++) {
    int p = 1LL * i * qp(n, P - 2) % P;
    int q = 1LL * (n - i) * qp(n, P - 2) % P * qp(K - 1, P - 2) % P;
    f[i] = (1LL * p * f[i - 1] + 1) % P * qp(q, P - 2) % P;
  }
  for (int i = n; i; i--) {
    s[i - 1] = (s[i] + f[i - 1]) % P;
  }
  m = cnt[-1];
  for (int i = 0; i <= m; i++) {
    p[i] = 1LL * binom(m, i) * qp(K - 1, m - i) % P * qp(qp(K, m), P - 2) % P;
  }
  int ans = 0, tot = K;
  for (auto &p : cnt)
    if (~p.first) {
      (ans += solve(p.second)) %= P, tot--;
    }
  ans = (ans + 1LL * tot * solve(0)) % P;
  ans = (ans - 1LL * (K - 1) * s[0] % P + P) % P;
  printf("%lld\n", 1LL * ans * qp(K - 1, P - 2) % P);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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 maxn = 100010, P = 1000000007;
int n, m, K, f[maxn], p[maxn], s[maxn], fact[maxn], finv[maxn];
unordered_map<int, int> cnt;
int qp(int x, int y) {
  int z = 1;
  for (; y; y >>= 1, x = 1LL * x * x % P) {
    if (y & 1) z = 1LL * z * x % P;
  }
  return z;
}
int binom(int n, int m) {
  return 1LL * fact[n] * finv[m] % P * finv[n - m] % P;
}
int solve(int x) {
  static unordered_map<int, int> mp;
  if (mp.count(x)) return mp[x];
  int ans = 0;
  for (int i = 0; i <= m; i++) ans = (ans + 1LL * p[i] * s[i + x]) % P;
  return mp[x] = ans;
}
int main() {
  scanf("%d %d", &n, &K);
  for (int i = fact[0] = finv[0] = 1; i <= n; i++) {
    fact[i] = 1LL * i * fact[i - 1] % P;
    finv[i] = qp(fact[i], P - 2);
  }
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x), cnt[x]++;
  }
  f[0] = K - 1;
  for (int i = 1; i < n; i++) {
    int p = 1LL * i * qp(n, P - 2) % P;
    int q = 1LL * (n - i) * qp(n, P - 2) % P * qp(K - 1, P - 2) % P;
    f[i] = (1LL * p * f[i - 1] + 1) % P * qp(q, P - 2) % P;
  }
  for (int i = n; i; i--) {
    s[i - 1] = (s[i] + f[i - 1]) % P;
  }
  m = cnt[-1];
  for (int i = 0; i <= m; i++) {
    p[i] = 1LL * binom(m, i) * qp(K - 1, m - i) % P * qp(qp(K, m), P - 2) % P;
  }
  int ans = 0, tot = K;
  for (auto &p : cnt)
    if (~p.first) {
      (ans += solve(p.second)) %= P, tot--;
    }
  ans = (ans + 1LL * tot * solve(0)) % P;
  ans = (ans - 1LL * (K - 1) * s[0] % P + P) % P;
  printf("%lld\n", 1LL * ans * qp(K - 1, P - 2) % P);
  return 0;
}

```