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;
int n, flag;
long long x, sum, ans, res;
long long a[100001];
long long pow(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1, a = (a * a) % 1000000007)
    if (b & 1) ans = (ans * a) % 1000000007;
  return ans;
}
int main() {
  scanf("%d%I64d", &n, &x);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    sum += a[i];
  }
  for (int i = 1; i <= n; i++) a[i] = (sum - a[i]);
  sort(a + 1, a + 1 + n);
  long long res = 0;
  flag = 1;
  for (int i = 1; i <= n; i++) {
    int j = i;
    while (a[j + 1] == a[i] && j <= n) j++;
    res += j - i + 1;
    if (j == n) {
      for (ans = a[j];; ans++) {
        if (res % x != 0) break;
        res /= x;
      }
      break;
    }
    for (long long ii = a[j]; ii < a[j + 1]; ii++) {
      if (!res) break;
      if (res % x != 0) {
        ans = ii;
        flag = 0;
        break;
      }
      res /= x;
    }
    if (!flag) break;
    i = j;
  }
  ans = pow(x, min(ans, sum));
  printf("%I64d\n", ans);
}
