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;
long long a[100005];
long long power(long long x, long long m) {
  long long ans = 1;
  x = x % 1000000007;
  while (m) {
    if (m & 1) ans = (ans * x) % 1000000007;
    x = (x * x) % 1000000007;
    m /= 2;
  }
  return ans % 1000000007;
}
int main() {
  long long n, x;
  long long s = 0;
  scanf("%I64d%I64d", &n, &x);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    s += a[i];
  }
  for (int i = 0; i < n; i++) {
    a[i] = s - a[i];
  }
  a[n] = -1;
  sort(a, a + n);
  long long cnt = 1, ans;
  for (int i = 1; i <= n; i++) {
    if (a[i] != a[i - 1]) {
      if (cnt % x == 0) {
        cnt /= x;
        a[i - 1] += 1;
        i--;
      } else {
        ans = a[i - 1];
        break;
      }
    } else {
      cnt++;
    }
  }
  ans = min(ans, s);
  printf("%I64d\n", power(x, ans));
}
