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 int a[100004];
long long int POW(long long int Base, long long int Exp) {
  long long int y, ret = 1;
  y = Base;
  while (Exp) {
    if (Exp & 1) ret = (ret * y) % 1000000007;
    y = (y * y) % 1000000007;
    Exp /= 2;
  }
  return ret % 1000000007;
}
int main() {
  long long int n, x, k, sum = 0, cnt = 0;
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 0; i < n; i++) a[i] = sum - a[i];
  sort(a, a + n);
  k = a[0];
  for (int i = 0; i < n;) {
    while (i < n && a[i] == k) {
      cnt++;
      i++;
    }
    if (cnt % x == 0) {
      k++;
      cnt /= x;
    } else
      break;
  }
  while (cnt % x == 0) {
    k++;
    cnt /= x;
  }
  k = min(k, sum);
  cout << POW(x, k) << endl;
  return 0;
}
