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 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;
  }
  