#include <bits/stdc++.h>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 100005;
const int mod = 1e9 + 7;
inline int qpow(int a, int b = mod - 2) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return res;
}
int fac[N], ifac[N];
inline void init(int n) {
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[n] = qpow(fac[n]);
  for (int i = n; i >= 1; i--) ifac[i - 1] = 1ll * ifac[i] * i % mod;
}
inline int C(int n, int m) {
  return 1ll * fac[n] * ifac[n - m] % mod * ifac[m] % mod;
}
map<int, int> cnt;
int f[N], zyk;
int a[N], buc[N], pw[N], n, k;
inline int gao(int x) {
  int res = 0;
  for (int i = 0; i <= zyk; i++) {
    int coef = 1ll * f[x + i] * C(zyk, i) % mod * pw[zyk - i] % mod;
    res = (res + coef) % mod;
  }
  return res;
}
int main() {
  init(N - 5);
  n = read(), k = read();
  pw[0] = 1;
  for (int i = 1; i <= n; i++) {
    pw[i] = 1ll * pw[i - 1] * (k - 1) % mod;
    a[i] = read();
    if (a[i] == -1)
      zyk++;
    else
      cnt[a[i]]++;
  }
  f[0] = 0, f[1] = 0;
  for (int a = 1; a <= n - 1; a++) {
    f[a + 1] = 1ll *
               (1ll * (k - 1) * a % mod * f[a - 1] +
                1ll * (2ll * a - 1ll * a * k % mod - n) * f[a] + 1ll * a * k) %
               mod * qpow(a - n) % mod;
    if (f[a + 1] < 0) f[a + 1] += mod;
  }
  int qwq = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] != -1 && cnt[a[i]]) {
      buc[cnt[a[i]]]++;
      cnt[a[i]] = 0;
      qwq++;
    }
  }
  buc[0] += k - qwq;
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    if (buc[i]) {
      ans = (ans + 1ll * buc[i] * gao(i)) % mod;
    }
  }
  ans = 1ll * ans * qpow(qpow(k, zyk)) % mod;
  print((ans + mod - f[n]) % mod, '\n');
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 100005;
const int mod = 1e9 + 7;
inline int qpow(int a, int b = mod - 2) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return res;
}
int fac[N], ifac[N];
inline void init(int n) {
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[n] = qpow(fac[n]);
  for (int i = n; i >= 1; i--) ifac[i - 1] = 1ll * ifac[i] * i % mod;
}
inline int C(int n, int m) {
  return 1ll * fac[n] * ifac[n - m] % mod * ifac[m] % mod;
}
map<int, int> cnt;
int f[N], zyk;
int a[N], buc[N], pw[N], n, k;
inline int gao(int x) {
  int res = 0;
  for (int i = 0; i <= zyk; i++) {
    int coef = 1ll * f[x + i] * C(zyk, i) % mod * pw[zyk - i] % mod;
    res = (res + coef) % mod;
  }
  return res;
}
int main() {
  init(N - 5);
  n = read(), k = read();
  pw[0] = 1;
  for (int i = 1; i <= n; i++) {
    pw[i] = 1ll * pw[i - 1] * (k - 1) % mod;
    a[i] = read();
    if (a[i] == -1)
      zyk++;
    else
      cnt[a[i]]++;
  }
  f[0] = 0, f[1] = 0;
  for (int a = 1; a <= n - 1; a++) {
    f[a + 1] = 1ll *
               (1ll * (k - 1) * a % mod * f[a - 1] +
                1ll * (2ll * a - 1ll * a * k % mod - n) * f[a] + 1ll * a * k) %
               mod * qpow(a - n) % mod;
    if (f[a + 1] < 0) f[a + 1] += mod;
  }
  int qwq = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] != -1 && cnt[a[i]]) {
      buc[cnt[a[i]]]++;
      cnt[a[i]] = 0;
      qwq++;
    }
  }
  buc[0] += k - qwq;
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    if (buc[i]) {
      ans = (ans + 1ll * buc[i] * gao(i)) % mod;
    }
  }
  ans = 1ll * ans * qpow(qpow(k, zyk)) % mod;
  print((ans + mod - f[n]) % mod, '\n');
  return 0;
}

```