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>
  constexpr int Maxn = 1e5 + 7;
  constexpr int Mod = 1e9 + 7;
  int fac[Maxn], ifac[Maxn];
  inline int fpm(int a, int b) {
    int res = 1;
    while (b) {
      if (b & 1) res = 1LL * res * a % Mod;
      a = 1LL * a * a % Mod;
      b >>= 1;
    }
    return res;
  }
  inline int C(int n, int m) {
    if (n < 0 || m < 0 || n < m) return 0;
    return 1LL * fac[n] * ifac[m] % Mod * ifac[n - m] % Mod;
  }
  std::unordered_map<int, int> cnt, mp;
  int n, m, k, f[Maxn], p[Maxn], s[Maxn];
  int solve(int x) {
    if (mp.count(x)) return mp[x];
    int res = 0;
    for (int i = 0; i <= m; ++i) res = (res + 1LL * p[i] * s[i + x] % Mod) % Mod;
    return mp[x] = res;
  }
  int main() {
    scanf("%d%d", &n, &k);
    fac[0] = 1;
    for (int i = 1; i <= n; ++i) fac[i] = 1LL * i * fac[i - 1] % Mod;
    ifac[n] = fpm(fac[n], Mod - 2);
    for (int i = n - 1; i >= 0; --i) ifac[i] = 1LL * (i + 1) * ifac[i + 1] % Mod;
    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 * fpm(n, Mod - 2) % Mod;
      int q = 1LL * (n - i) * fpm(n, Mod - 2) % Mod * fpm(k - 1, Mod - 2) % Mod;
      f[i] = (1LL * p * f[i - 1] % Mod + 1) * fpm(q, Mod - 2) % Mod;
    }
    for (int i = n; i; --i) s[i - 1] = (s[i] + f[i - 1]) % Mod;
    m = cnt[-1];
    for (int i = 0; i <= m; ++i)
      p[i] =
          1LL * C(m, i) * fpm(k - 1, m - i) % Mod * fpm(fpm(k, m), Mod - 2) % Mod;
    int res = 0, tot = k;
    for (auto x : cnt)
      if (x.first != -1) {
        res = (res + solve(x.second)) % Mod;
        --tot;
      }
    res = (res + 1LL * tot * solve(0) % Mod) % Mod;
    res = (res - 1LL * (k - 1) * s[0] % Mod + Mod) % Mod;
    printf("%d\n", 1LL * res * fpm(k - 1, Mod - 2) % Mod);
  }
  