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;
  constexpr long long mod = 1000000007;
  const long long INF = mod * mod;
  const long double eps = 1e-8;
  const long double pi = acosl(-1.0);
  long long mod_pow(long long x, long long n, long long m = mod) {
    if (n < 0) {
      long long res = mod_pow(x, -n, m);
      return mod_pow(res, m - 2, m);
    }
    if (abs(x) >= m) x %= m;
    if (x < 0) x += m;
    long long res = 1;
    while (n) {
      if (n & 1) res = res * x % m;
      x = x * x % m;
      n >>= 1;
    }
    return res;
  }
  struct modint {
    long long n;
    modint() : n(0) { ; }
    modint(long long m) : n(m) {
      if (n >= mod)
        n %= mod;
      else if (n < 0)
        n = (n % mod + mod) % mod;
    }
    operator int() { return n; }
  };
  bool operator==(modint a, modint b) { return a.n == b.n; }
  modint operator+=(modint& a, modint b) {
    a.n += b.n;
    if (a.n >= mod) a.n -= mod;
    return a;
  }
  modint operator-=(modint& a, modint b) {
    a.n -= b.n;
    if (a.n < 0) a.n += mod;
    return a;
  }
  modint operator*=(modint& a, modint b) {
    a.n = ((long long)a.n * b.n) % mod;
    return a;
  }
  modint operator+(modint a, modint b) { return a += b; }
  modint operator-(modint a, modint b) { return a -= b; }
  modint operator*(modint a, modint b) { return a *= b; }
  modint operator^(modint a, long long n) {
    if (n == 0) return modint(1);
    modint res = (a * a) ^ (n / 2);
    if (n % 2) res = res * a;
    return res;
  }
  long long inv(long long a, long long p) {
    return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
  }
  modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }
  modint operator/=(modint& a, modint b) {
    a = a / b;
    return a;
  }
  const int max_n = 1 << 18;
  modint fact[max_n], factinv[max_n];
  void init_f() {
    fact[0] = modint(1);
    for (int i = 0; i < max_n - 1; i++) {
      fact[i + 1] = fact[i] * modint(i + 1);
    }
    factinv[max_n - 1] = modint(1) / fact[max_n - 1];
    for (int i = max_n - 2; i >= 0; i--) {
      factinv[i] = factinv[i + 1] * modint(i + 1);
    }
  }
  modint comb(int a, int b) {
    if (a < 0 || b < 0 || a < b) return 0;
    return fact[a] * factinv[b] * factinv[a - b];
  }
  modint combP(int a, int b) {
    if (a < 0 || b < 0 || a < b) return 0;
    return fact[a] * factinv[a - b];
  }
  using mP = pair<modint, modint>;
  void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    sort((a).begin(), (a).end());
    vector<mP> memo(n);
    memo[0] = {1, 0};
    for (int c = 0; c <= n - 2; c++) {
      modint A = (long long)n * k - c - (long long)(n - c) * (k - 1);
      modint B = -(long long)c * (k - 1);
      modint x = A * memo[c].first;
      if (c > 0) x += B * memo[c - 1].first;
      modint y = A * memo[c].second;
      if (c > 0) y += B * memo[c - 1].second;
      y -= (long long)n * k;
      x /= n - c;
      y /= n - c;
      memo[c + 1] = {x, y};
    }
    int c = n - 1;
    modint A = (long long)n * k - c - (long long)(n - c) * (k - 1);
    modint B = -(long long)c * (k - 1);
    modint x = A * memo[c].first;
    if (c > 0) x += B * memo[c - 1].first;
    modint y = A * memo[c].second;
    if (c > 0) y += B * memo[c - 1].second;
    y -= (long long)n * k;
    vector<modint> ex(n);
    ex[0] = (modint)-y / x;
    for (int c = 1; c < n; c++) {
      ex[c] = memo[c].first * ex[0] + memo[c].second;
    }
    int ad = 0;
    vector<int> cs;
    int rest = k;
    for (int i = 0; i < n; i++) {
      int le = i;
      while (i + 1 < n && a[i] == a[i + 1]) i++;
      int len = i - le + 1;
      if (a[i] < 0) {
        ad = len;
      } else {
        rest--;
        cs.push_back(len);
      }
    }
    sort((cs).begin(), (cs).end());
    vector<pair<int, int> > ps;
    for (int i = 0; i < cs.size(); i++) {
      int le = i;
      while (i + 1 < cs.size() && cs[i] == cs[i + 1]) i++;
      int cnt = i - le + 1;
      ps.push_back({cs[i], cnt});
    }
    ps.push_back({0, rest});
    modint ans = 0;
    vector<modint> coef(ad + 1);
    for (int i = 0; i < ad + 1; i++) {
      coef[i] = comb(ad, i) * (modint)mod_pow(k - 1, ad - i);
    }
    ex.push_back(0);
    for (pair<int, int> p : ps) {
      for (int i = 0; i <= ad; i++) {
        modint val = ex[p.first + i];
        val *= p.second;
        ans += val * coef[i];
      }
    }
    ans -= (modint)(k - 1) * ex[0] * (modint)mod_pow(k, ad);
    ans /= k;
    ans /= mod_pow(k, ad);
    cout << ans << "\n";
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    init_f();
    solve();
    return 0;
  }
  