#include <bits/stdc++.h>
using namespace std;
const int N = 110000;
const int M = 1100000;
const long long mod = 1e9 + 7;
const int inf = (int)1e9;
const double eps = 1e-9;
long long n, k, a[N];
long long f[N], inv[N], fac[N], sinv[N], g[N];
long long pw[N];
long long pow_mod(long long a, long long e) {
  long long res = 1;
  for (; e; a = a * a % mod, e >>= 1)
    if (e & 1) res = res * a % mod;
  return res;
}
long long num;
long long C(int n, int m) {
  long long res = fac[n] * sinv[m] % mod;
  res = res * sinv[n - m] % mod;
  return res;
}
map<int, int> S;
long long cal(int i) {
  if (g[i] != -1) return g[i];
  long long fm = pow_mod(pow_mod(k, num), mod - 2);
  long long res = 0;
  for (int j = 0; j <= num; j++) {
    long long tmp = C(num, j) * f[i + j] % mod;
    res = (res + tmp * pw[num - j]) % mod;
  }
  res = res * fm % mod;
  g[i] = res;
  return g[i];
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == -1)
      num++;
    else
      S[a[i]]++;
  }
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; i++) inv[i] = pow_mod(i, mod - 2);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * (long long)i) % mod;
  sinv[0] = 1;
  for (int i = 1; i <= n; i++) sinv[i] = (sinv[i - 1] * inv[i]) % mod;
  f[0] = 0, f[1] = 0;
  for (int i = 2; i <= n; i++) {
    long long a = i - 1;
    long long tmp1 = (k - 1) * a + n - a;
    tmp1 %= mod;
    tmp1 = tmp1 * f[i - 1] % mod;
    long long tmp2 = (k - 1) * a % mod;
    tmp2 %= mod;
    tmp2 = tmp2 * f[i - 2] % mod;
    long long tmp3 = a * k % mod;
    long long tmp = (tmp1 - tmp2 - tmp3 + mod + mod) % mod;
    f[i] = tmp * inv[n - a] % mod;
  }
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = (pw[i - 1] * (k - 1)) % mod;
  memset(g, -1, sizeof(g));
  long long ans = 0;
  for (auto p : S) {
    int val = p.second;
    ans = (ans + cal(val)) % mod;
  }
  ans = (ans + (k - S.size()) * cal(0)) % mod;
  cout << (ans - f[n] + mod) % mod << "\n";
  return 0;
}

### Prompt

Your task is to create 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 N = 110000;
const int M = 1100000;
const long long mod = 1e9 + 7;
const int inf = (int)1e9;
const double eps = 1e-9;
long long n, k, a[N];
long long f[N], inv[N], fac[N], sinv[N], g[N];
long long pw[N];
long long pow_mod(long long a, long long e) {
  long long res = 1;
  for (; e; a = a * a % mod, e >>= 1)
    if (e & 1) res = res * a % mod;
  return res;
}
long long num;
long long C(int n, int m) {
  long long res = fac[n] * sinv[m] % mod;
  res = res * sinv[n - m] % mod;
  return res;
}
map<int, int> S;
long long cal(int i) {
  if (g[i] != -1) return g[i];
  long long fm = pow_mod(pow_mod(k, num), mod - 2);
  long long res = 0;
  for (int j = 0; j <= num; j++) {
    long long tmp = C(num, j) * f[i + j] % mod;
    res = (res + tmp * pw[num - j]) % mod;
  }
  res = res * fm % mod;
  g[i] = res;
  return g[i];
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == -1)
      num++;
    else
      S[a[i]]++;
  }
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; i++) inv[i] = pow_mod(i, mod - 2);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * (long long)i) % mod;
  sinv[0] = 1;
  for (int i = 1; i <= n; i++) sinv[i] = (sinv[i - 1] * inv[i]) % mod;
  f[0] = 0, f[1] = 0;
  for (int i = 2; i <= n; i++) {
    long long a = i - 1;
    long long tmp1 = (k - 1) * a + n - a;
    tmp1 %= mod;
    tmp1 = tmp1 * f[i - 1] % mod;
    long long tmp2 = (k - 1) * a % mod;
    tmp2 %= mod;
    tmp2 = tmp2 * f[i - 2] % mod;
    long long tmp3 = a * k % mod;
    long long tmp = (tmp1 - tmp2 - tmp3 + mod + mod) % mod;
    f[i] = tmp * inv[n - a] % mod;
  }
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = (pw[i - 1] * (k - 1)) % mod;
  memset(g, -1, sizeof(g));
  long long ans = 0;
  for (auto p : S) {
    int val = p.second;
    ans = (ans + cal(val)) % mod;
  }
  ans = (ans + (k - S.size()) * cal(0)) % mod;
  cout << (ans - f[n] + mod) % mod << "\n";
  return 0;
}

```