description: 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
solution: #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;
  }
  