#include <bits/stdc++.h>
const int P = 1000000007, Q = P - 1;
int mpow(int x, int k) {
  if (k == 0) return 1;
  int ret = mpow(x * (unsigned long long)x % P, k >> 1);
  if (k & 1) ret = ret * (unsigned long long)x % P;
  return ret;
}
const int N = 100010;
int e[N], fac[N], ifac[N], inv[N], pp[N], pq[N];
int cnt[N];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int n, k;
  std::cin >> n >> k;
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * (unsigned long long)i % P;
  ifac[n] = mpow(fac[n], P - 2);
  for (int i = n; i; --i) ifac[i - 1] = ifac[i] * (unsigned long long)i % P;
  for (int i = 1; i <= n; ++i)
    inv[i] = ifac[i] * (unsigned long long)fac[i - 1] % P;
  int p = mpow(k, P - 2), in = inv[n];
  e[0] = k;
  for (int i = 1; i != n; ++i) {
    e[i] =
        (i * (unsigned long long)(k - 1) % P * p % P * in % P * e[i - 1] + 1) %
        P;
    e[i] = e[i] * (unsigned long long)n % P * k % P * inv[n - i] % P;
  }
  for (int i = n - 1; i >= 0; --i) e[i] = (e[i] + e[i + 1]) % P;
  std::unordered_map<int, int> mp;
  int oth = 0;
  for (int rep = 0; rep != n; ++rep) {
    int ai;
    std::cin >> ai;
    if (ai == -1)
      ++oth;
    else
      ++mp[ai];
  }
  int q = 1 - p;
  if (q < 0) q += P;
  pp[0] = 1;
  for (int i = 1; i <= n; ++i) pp[i] = pp[i - 1] * (unsigned long long)p % P;
  pq[0] = 1;
  for (int i = 1; i <= n; ++i) pq[i] = pq[i - 1] * (unsigned long long)q % P;
  auto cal = [&](int c) {
    int ret = 0;
    for (int i = 0; i <= oth; ++i) {
      ret = (ret + pp[i] * (unsigned long long)pq[oth - i] % P * ifac[i] % P *
                       ifac[oth - i] % P * e[c + i]) %
            P;
    }
    return ret * (unsigned long long)fac[oth] % P;
  };
  cnt[0] = k;
  for (const auto& pr : mp) {
    ++cnt[pr.second];
    --cnt[0];
  }
  int ans = (P - k + 1) * (unsigned long long)e[0] % P;
  for (int i = 0; i <= n; ++i)
    if (cnt[i]) ans = (ans + cnt[i] * (unsigned long long)cal(i)) % P;
  ans = ans * (unsigned long long)p % P;
  std::cout << ans << '\n';
  return 0;
}

### Prompt

Generate 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>
const int P = 1000000007, Q = P - 1;
int mpow(int x, int k) {
  if (k == 0) return 1;
  int ret = mpow(x * (unsigned long long)x % P, k >> 1);
  if (k & 1) ret = ret * (unsigned long long)x % P;
  return ret;
}
const int N = 100010;
int e[N], fac[N], ifac[N], inv[N], pp[N], pq[N];
int cnt[N];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int n, k;
  std::cin >> n >> k;
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * (unsigned long long)i % P;
  ifac[n] = mpow(fac[n], P - 2);
  for (int i = n; i; --i) ifac[i - 1] = ifac[i] * (unsigned long long)i % P;
  for (int i = 1; i <= n; ++i)
    inv[i] = ifac[i] * (unsigned long long)fac[i - 1] % P;
  int p = mpow(k, P - 2), in = inv[n];
  e[0] = k;
  for (int i = 1; i != n; ++i) {
    e[i] =
        (i * (unsigned long long)(k - 1) % P * p % P * in % P * e[i - 1] + 1) %
        P;
    e[i] = e[i] * (unsigned long long)n % P * k % P * inv[n - i] % P;
  }
  for (int i = n - 1; i >= 0; --i) e[i] = (e[i] + e[i + 1]) % P;
  std::unordered_map<int, int> mp;
  int oth = 0;
  for (int rep = 0; rep != n; ++rep) {
    int ai;
    std::cin >> ai;
    if (ai == -1)
      ++oth;
    else
      ++mp[ai];
  }
  int q = 1 - p;
  if (q < 0) q += P;
  pp[0] = 1;
  for (int i = 1; i <= n; ++i) pp[i] = pp[i - 1] * (unsigned long long)p % P;
  pq[0] = 1;
  for (int i = 1; i <= n; ++i) pq[i] = pq[i - 1] * (unsigned long long)q % P;
  auto cal = [&](int c) {
    int ret = 0;
    for (int i = 0; i <= oth; ++i) {
      ret = (ret + pp[i] * (unsigned long long)pq[oth - i] % P * ifac[i] % P *
                       ifac[oth - i] % P * e[c + i]) %
            P;
    }
    return ret * (unsigned long long)fac[oth] % P;
  };
  cnt[0] = k;
  for (const auto& pr : mp) {
    ++cnt[pr.second];
    --cnt[0];
  }
  int ans = (P - k + 1) * (unsigned long long)e[0] % P;
  for (int i = 0; i <= n; ++i)
    if (cnt[i]) ans = (ans + cnt[i] * (unsigned long long)cal(i)) % P;
  ans = ans * (unsigned long long)p % P;
  std::cout << ans << '\n';
  return 0;
}

```