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 double pi = acos(-1.0);
int powermod(int a, long long p, int mod) {
  long long t = a;
  int r = 1;
  while (p) {
    if (p & 1) r = ((long long)r * t) % mod;
    t = (t * t) % mod;
    p >>= 1;
  }
  return r;
}
int n, x;
int a[100000];
int cnt[2 * 100000 + 1];
int main() {
  scanf("%d", &n);
  scanf("%d", &x);
  int mina = 1000000000;
  int maxa = -1;
  long long suma = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    mina = min(a[i], mina);
    maxa = max(a[i], maxa);
    suma += a[i];
  }
  long long num = suma - maxa;
  if (n >= x) {
    for (int i = 0; i < n; i++) {
      if (2LL * (maxa - a[i]) <= 2 * n) {
        cnt[maxa - a[i]]++;
      }
    }
    int p = 0;
    while (cnt[p] > 0 && cnt[p] % x == 0) {
      cnt[p + 1] += cnt[p] / x;
      p++;
    }
    num += p;
  }
  printf("%d\n", powermod(x, min(num, suma), 1000000007));
  return 0;
}
