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;
  template <typename T>
  inline void read(T &f) {
    f = 0;
    T fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') {
        fu = -1;
      }
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      f = (f << 3) + (f << 1) + (c & 15);
      c = getchar();
    }
    f *= fu;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10)
      putchar(x + 48);
    else
      print(x / 10), putchar(x % 10 + 48);
  }
  template <typename T>
  void print(T x, char t) {
    print(x);
    putchar(t);
  }
  const int md = 1e9 + 7;
  inline int add(int x, int y) {
    if (x + y >= md) return x + y - md;
    return x + y;
  }
  inline void addx(int &x, int y) {
    x += y;
    if (x >= md) x -= md;
  }
  inline int sub(int x, int y) {
    if (x < y) return x - y + md;
    return x - y;
  }
  inline void subx(int &x, int y) {
    x -= y;
    if (x < 0) x += md;
  }
  inline int mul(int x, int y) { return 1ull * x * y % md; }
  inline int fpow(int x, int y) {
    int ans = 1;
    while (y) {
      if (y & 1) ans = mul(ans, x);
      y >>= 1;
      x = mul(x, x);
    }
    return ans;
  }
  const int N = 1e5 + 5;
  int a[N], f[N], fac[N], inv[N], t[N], tt[N], pw[N];
  int n, k, ans;
  inline int C(int n, int m) {
    if (n < m || m < 0) return 0;
    return mul(fac[n], mul(inv[m], inv[n - m]));
  }
  int main() {
    fac[0] = 1;
    for (int i = 1; i < N; i++) fac[i] = mul(fac[i - 1], i);
    inv[N - 1] = fpow(fac[N - 1], md - 2);
    for (int i = N - 1; i >= 1; i--) inv[i - 1] = mul(inv[i], i);
    read(n);
    read(k);
    pw[0] = 1;
    for (int i = 1; i <= n; i++) pw[i] = mul(pw[i - 1], k - 1);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 0; i < n; i++) {
      int tem = fpow(n - i, md - 2);
      int c2 = mul(sub(0, mul(i, k - 1)), tem);
      int c1 = mul(sub(mul(n, k), add(i, mul(n - i, k - 1))), tem);
      int c0 = mul(sub(0, mul(k, i)), tem);
      f[i + 1] = add(c0, mul(c1, f[i]));
      if (i != 0) f[i + 1] = add(f[i + 1], mul(c2, f[i - 1]));
    }
    sort(a + 1, a + n + 1);
    int cnt = 0;
    t[0] = k;
    for (int l = 1, r; l <= n; l = r + 1) {
      r = l;
      while (r < n && a[r + 1] == a[l]) ++r;
      if (a[l] == -1) {
        cnt = r - l + 1;
        continue;
      }
      ++t[r - l + 1];
      --t[0];
    }
    for (int i = 0; i <= n; i++) {
      if (t[i]) {
        for (int j = 0; j <= cnt; j++) {
          tt[i + j] = add(tt[i + j], mul(t[i], mul(C(cnt, j), pw[cnt - j])));
        }
      }
    }
    for (int i = 0; i <= n; i++) {
      ans = add(ans, mul(tt[i], f[i]));
    }
    ans = mul(ans, fpow(fpow(k, cnt), md - 2));
    print(sub(ans, f[n]), '\n');
    return 0;
  }
  