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 = 1e6 + 11, mod = 1e9 + 7;
  int f[N], g[N], a[N], b[N], c[N], d[N], fac[N], ifac[N], ik[N], ik0[N];
  int n, s, ans, cnt, k, m;
  int fp(int a, int b) {
    int res = 1;
    for (; b; b >>= 1, a = 1ll * a * a % mod)
      if (b & 1) res = 1ll * res * a % mod;
    return res;
  }
  int C(int n, int m) { return 1ll * fac[n] * ifac[m] % mod * ifac[n - m] % mod; }
  int main() {
    scanf("%d%d", &n, &k);
    fac[0] = 1;
    for (int i = (1); i <= (n); i++) fac[i] = 1ll * fac[i - 1] * i % mod;
    for (int i = (0); i <= (n); i++) ifac[i] = fp(fac[i], mod - 2);
    ik[0] = ik0[0] = 1;
    for (int i = (1); i <= (n); i++) {
      ik[i] = 1ll * ik[i - 1] * fp(k, mod - 2) % mod,
      ik0[i] = 1ll * ik0[i - 1] * (1 - fp(k, mod - 2)) % mod;
    }
    s = n;
    for (int i = (1); i <= (n); i++) {
      scanf("%d", a + i);
      if (a[i] == -1) {
        --i;
        --n;
        ++cnt;
        continue;
      }
      b[i] = a[i];
    }
    sort(b + 1, b + n + 1);
    m = unique(b + 1, b + n + 1) - b - 1;
    int p = 0;
    for (int i = (1); i <= (n); i++) {
      a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
      if (!c[a[i]]) ++p;
      ++c[a[i]];
    }
    n = k;
    g[0] = n;
    for (int i = (1); i <= (s - 1); i++)
      g[i] = 1ll * fp(s - i, mod - 2) *
             (1ll * s * n % mod + 1ll * i * (n - 1) % mod * g[i - 1] % mod) % mod;
    f[s] = 0;
    for (int i = s - 1; ~i; --i) f[i] = (f[i + 1] + g[i]) % mod;
    for (int i = (1); i <= (p); i++) ++d[c[i]];
    d[0] = n - m;
    for (int i = (0); i <= (s); i++) {
      if (d[i] > 0) {
        for (int j = (0); j <= (cnt); j++)
          ans = (ans + 1ll * d[i] * f[i + j] % mod * C(cnt, j) % mod * ik[j] %
                           mod * ik0[cnt - j]) %
                mod;
      }
    }
    ans = 1ll * (ans - 1ll * f[0] * (n - 1)) % mod * fp(n, mod - 2) % mod;
    ans = (ans + mod) % mod;
    printf("%d\n", ans);
    return 0;
  }
  