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 long long M = 1000000007;
  long long n, k;
  long long a[100050];
  long long cnt[100050];
  map<long long, long long> dict;
  long long f[100050];
  long long fac[100000 + 50], ifac[100000 + 50];
  long long pwk[100050];
  long long ksm(long long a, long long x = M - 2) {
    long long ret = 1;
    for (; x; x >>= 1, a = a * a % M)
      if (x & 1) ret = ret * a % M;
    return ret;
  }
  long long C(long long n, long long m) {
    if (m < 0 || m > n) return 0;
    return fac[n] * ifac[m] % M * ifac[n - m] % M;
  }
  void prepare() {
    fac[0] = 1;
    for (int i = 1; i <= 100000; i++) fac[i] = fac[i - 1] * i % M;
    ifac[100000] = ksm(fac[100000]);
    for (int i = 100000 - 1; i >= 0; i--) ifac[i] = ifac[i + 1] * (i + 1) % M;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    prepare();
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      dict[a[i]]++;
    }
    cnt[0] = k;
    for (auto y : dict)
      if (y.first != -1) cnt[y.second]++, cnt[0]--;
    int r = 0;
    if (dict.find(-1) != dict.end()) r = dict[-1];
    for (int i = 1; i < n; i++) {
      f[i + 1] = ksm(M + i - n) *
                 ((k - 1) * i % M * f[i - 1] % M +
                  ((M + 2 - k) * i % M - n + M) * f[i] % M + i * k % M) %
                 M;
    }
    long long invpwkr = ksm(ksm(k, r));
    pwk[0] = 1;
    for (int i = 1; i <= n; i++) pwk[i] = pwk[i - 1] * (k - 1) % M;
    long long ans = (M - (f[0] * (k - 1) + f[n]) % M) % M;
    for (int i = 0; i <= n; i++)
      if (cnt[i] != 0) {
        for (int j = 0; j <= r; j++)
          ans = (ans +
                 C(r, j) * f[i + j] % M * cnt[i] % M * invpwkr % M * pwk[r - j]) %
                M;
      }
    cout << ans << '\n';
    return 0;
  }
  