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 pri = 1000000007;
int ksm(int x, int y) {
  if (!y) return 1;
  int t = ksm(x, y / 2);
  if (y & 1) return (long long)t * t % pri * x % pri;
  return (long long)t * t % pri;
}
bool check(int x, int y, int z) {
  while (z--) {
    if (x % y) return 0;
    x /= y;
  }
  return 1;
}
int a[100005];
int main() {
  int n, m;
  long long sum = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), sum += a[i];
  sort(a + 1, a + 1 + n);
  int d = 0;
  long long now = sum - a[n];
  for (int i = n; i; i--) {
    while (d % m == 0 && now < sum - a[i]) d /= m, now++;
    if (sum - a[i] == now)
      d++;
    else {
      printf("%d", ksm(m, now % (pri - 1)));
      return 0;
    }
  }
  while (d % m == 0) d /= m, now++;
  printf("%d", ksm(m, min(now, sum) % (pri - 1)));
  return 0;
}
