#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
const int maxn = 1e5 + 10;
map<int, int> A, B;
long long k, f[maxn], p[maxn], jc[maxn], s[maxn];
int n, m;
long long pow_(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % M;
    x = x * x % M;
    y >>= 1;
  }
  return ret;
}
long long inv(long long x) { return pow_(x, M - 2); }
long long C(int x, int y) { return jc[x] * inv(jc[y] * jc[x - y] % M) % M; }
long long solve(int x) {
  if (B.find(x) != B.end()) return B[x];
  long long ret = 0;
  for (int i = 0; i <= m; i++) ret = (ret + p[i] * s[x + i]) % M;
  return B[x] = ret;
}
int main() {
  cin >> n >> k;
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % M;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    A[x]++;
  }
  f[0] = k - 1;
  for (int i = 1; i < n; i++) {
    long long p = inv(n) * i % M;
    long long q = inv(n) * (n - i) % M * inv(k - 1) % M;
    f[i] = (p * f[i - 1] + 1) % M * inv(q) % M;
  }
  s[n] = 0;
  for (int i = n; i; i--) s[i - 1] = (s[i] + f[i - 1]) % M;
  m = A[-1];
  for (int i = 0; i <= m; i++)
    p[i] = C(m, i) * pow_(k - 1, m - i) % M * inv(pow_(k, m)) % M;
  long long tot = k;
  long long ans = 0;
  for (auto x : A) {
    if (x.first == -1) continue;
    ans = (ans + solve(x.second)) % M;
    tot--;
  }
  ans = (ans + tot * solve(0)) % M;
  ans = (ans - (k - 1) * s[0]) % M;
  ans = ans * inv(k - 1) % M;
  cout << (ans + M) % M << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 M = 1000000007;
const int maxn = 1e5 + 10;
map<int, int> A, B;
long long k, f[maxn], p[maxn], jc[maxn], s[maxn];
int n, m;
long long pow_(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % M;
    x = x * x % M;
    y >>= 1;
  }
  return ret;
}
long long inv(long long x) { return pow_(x, M - 2); }
long long C(int x, int y) { return jc[x] * inv(jc[y] * jc[x - y] % M) % M; }
long long solve(int x) {
  if (B.find(x) != B.end()) return B[x];
  long long ret = 0;
  for (int i = 0; i <= m; i++) ret = (ret + p[i] * s[x + i]) % M;
  return B[x] = ret;
}
int main() {
  cin >> n >> k;
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % M;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    A[x]++;
  }
  f[0] = k - 1;
  for (int i = 1; i < n; i++) {
    long long p = inv(n) * i % M;
    long long q = inv(n) * (n - i) % M * inv(k - 1) % M;
    f[i] = (p * f[i - 1] + 1) % M * inv(q) % M;
  }
  s[n] = 0;
  for (int i = n; i; i--) s[i - 1] = (s[i] + f[i - 1]) % M;
  m = A[-1];
  for (int i = 0; i <= m; i++)
    p[i] = C(m, i) * pow_(k - 1, m - i) % M * inv(pow_(k, m)) % M;
  long long tot = k;
  long long ans = 0;
  for (auto x : A) {
    if (x.first == -1) continue;
    ans = (ans + solve(x.second)) % M;
    tot--;
  }
  ans = (ans + tot * solve(0)) % M;
  ans = (ans - (k - 1) * s[0]) % M;
  ans = ans * inv(k - 1) % M;
  cout << (ans + M) % M << endl;
}

```