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 N = 110000;
  const int M = 1100000;
  const long long mod = 1e9 + 7;
  const int inf = (int)1e9;
  const double eps = 1e-9;
  long long n, k, a[N];
  long long f[N], inv[N], fac[N], sinv[N], g[N];
  long long pw[N];
  long long pow_mod(long long a, long long e) {
    long long res = 1;
    for (; e; a = a * a % mod, e >>= 1)
      if (e & 1) res = res * a % mod;
    return res;
  }
  long long num;
  long long C(int n, int m) {
    long long res = fac[n] * sinv[m] % mod;
    res = res * sinv[n - m] % mod;
    return res;
  }
  map<int, int> S;
  long long cal(int i) {
    if (g[i] != -1) return g[i];
    long long fm = pow_mod(pow_mod(k, num), mod - 2);
    long long res = 0;
    for (int j = 0; j <= num; j++) {
      long long tmp = C(num, j) * f[i + j] % mod;
      res = (res + tmp * pw[num - j]) % mod;
    }
    res = res * fm % mod;
    g[i] = res;
    return g[i];
  }
  int main() {
    cin.sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (a[i] == -1)
        num++;
      else
        S[a[i]]++;
    }
    inv[0] = inv[1] = 1;
    for (int i = 2; i <= n; i++) inv[i] = pow_mod(i, mod - 2);
    fac[0] = 1;
    for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * (long long)i) % mod;
    sinv[0] = 1;
    for (int i = 1; i <= n; i++) sinv[i] = (sinv[i - 1] * inv[i]) % mod;
    f[0] = 0, f[1] = 0;
    for (int i = 2; i <= n; i++) {
      long long a = i - 1;
      long long tmp1 = (k - 1) * a + n - a;
      tmp1 %= mod;
      tmp1 = tmp1 * f[i - 1] % mod;
      long long tmp2 = (k - 1) * a % mod;
      tmp2 %= mod;
      tmp2 = tmp2 * f[i - 2] % mod;
      long long tmp3 = a * k % mod;
      long long tmp = (tmp1 - tmp2 - tmp3 + mod + mod) % mod;
      f[i] = tmp * inv[n - a] % mod;
    }
    pw[0] = 1;
    for (int i = 1; i <= n; i++) pw[i] = (pw[i - 1] * (k - 1)) % mod;
    memset(g, -1, sizeof(g));
    long long ans = 0;
    for (auto p : S) {
      int val = p.second;
      ans = (ans + cal(val)) % mod;
    }
    ans = (ans + (k - S.size()) * cal(0)) % mod;
    cout << (ans - f[n] + mod) % mod << "\n";
    return 0;
  }
  