Simon has a prime number x and an array of non-negative integers a1, a2, ..., an.

Simon loves fractions very much. Today he wrote out number <image> on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a fraction: <image>, where number t equals xa1 + a2 + ... + an. Now Simon wants to reduce the resulting fraction. 

Help him, find the greatest common divisor of numbers s and t. As GCD can be rather large, print it as a remainder after dividing it by number 1000000007 (109 + 7).

Input

The first line contains two positive integers n and x (1 ≤ n ≤ 105, 2 ≤ x ≤ 109) — the size of the array and the prime number.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ a1 ≤ a2 ≤ ... ≤ an ≤ 109). 

Output

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

2 2
2 2


Output

8


Input

3 3
1 2 3


Output

27


Input

2 2
29 29


Output

73741817


Input

4 5
0 0 0 0


Output

1

Note

In the first sample <image>. Thus, the answer to the problem is 8.

In the second sample, <image>. The answer to the problem is 27, as 351 = 13·27, 729 = 27·27.

In the third sample the answer to the problem is 1073741824 mod 1000000007 = 73741817.

In the fourth sample <image>. Thus, the answer to the problem is 1.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
const int maxm = 1e3 + 5;
const long long mod = 1e9 + 7;
long long quickpow(long long n, long long k, long long mod) {
  long long ans = 1;
  n = n % mod;
  while (k) {
    if (k & 1) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k /= 2;
    ans %= mod;
  }
  return ans % mod;
}
long long n, m, p;
long long a[maxn], b[maxn];
int main() {
  while (~scanf("%lld %lld", &n, &p)) {
    long long sum = 0;
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      sum += a[i];
    }
    for (int i = 0; i < n; i++) {
      b[i] = sum - a[i];
    }
    sort(b, b + n);
    long long cnt = 1;
    b[n] = -1;
    long long minn;
    for (int i = 1; i <= n; i++) {
      if (b[i] == b[i - 1])
        cnt++;
      else {
        if (cnt % p) {
          minn = b[i - 1];
          break;
        } else {
          b[i - 1]++;
          cnt /= p;
          i--;
        }
      }
    }
    long long ans = min(minn, sum);
    printf("%lld\n", quickpow(p, ans, mod));
  }
  return 0;
}
