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
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr LL mod = 1'000'000'007;
constexpr int maxn = 100000 + 1;
LL power(LL a, LL r) {
    LL res = 1;
    for (; r; r >>= 1, a = a * a % mod)
        if (r & 1) res = res * a % mod;
    return res;
}
LL g[maxn], f[maxn];
LL F[maxn], G[maxn];
LL S[maxn];
LL C(LL n, LL m) {
    return F[n] * G[m] % mod * G[n - m] % mod;
}
int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    map<int, int> mp;
    for (int i = 0; i <= n; i += 1) G[i] = power(F[i] = i ? F[i - 1] * i % mod : 1, mod - 2);
    for (int i = 0, a; i < n; i += 1) {
        cin >> a;
        mp[a] += 1;
    }
    for (int i = 0; i < n; i += 1) 
        g[i] = i ? (i * g[i - 1] % mod * (k - 1) + (LL)n * k) % mod * power(n - i, mod - 2) % mod : k;
    for (int i = n - 1; i >= 0; i -= 1)
        f[i] = g[i] + f[i + 1];
    map<int, int> count;
    count[0] = k;
    for (auto [x, y] : mp) if (x != -1) {
        count[y] += 1;
        count[0] -= 1;
    }
    int r = mp[-1];
    for (auto [x, y] : count) {
        //cout << x << " " << y << "\n";
        for (int i = 0; i <= r; i += 1) S[i] = (S[i] + y * f[x + i]) % mod;
    }
    LL ans = 0;
    for (int i = 0; i <= r; i += 1) {
        LL p = C(r, i) * power(k - 1, r - i) % mod * power(power(k, mod - 2), r) % mod;
        ans = (ans + p * S[i]) % mod;
    }
    cout << (ans + mod - f[0] * (k - 1) % mod) % mod * power(k, mod - 2) % mod;
    return 0;
}