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;
unsigned long long re(unsigned long long a, unsigned long long b) {
  if (b == 0) return 1;
  if (b % 2)
    return (re(a, b - 1) * a) % 1000000007;
  else
    return re((a * a) % 1000000007, b / 2);
}
int main() {
  unsigned long long n, x;
  cin >> n >> x;
  vector<unsigned long long> arr;
  unsigned long long sum = 0;
  unsigned long long powa = 0;
  for (int i = 0; i < n; i++) {
    unsigned long long v;
    cin >> v;
    sum += v;
    arr.push_back(v);
  }
  for (int i = 0; i < n; i++) {
    arr[i] = (sum - arr[i]);
  }
  sort(arr.begin(), arr.end());
  reverse(arr.begin(), arr.end());
  int coun = 0;
  for (int i = arr.size() - 1; i >= 0; i--) {
    if (arr[i] != arr[i - 1]) {
      coun++;
      int k = coun;
      unsigned long long l = arr[i];
      if (coun % x == 0 && coun > 0) {
        while (k--) {
          arr.pop_back();
        }
        coun /= x;
        while (coun--) {
          arr.push_back(l + 1);
        }
        i = arr.size();
        powa = arr[i];
        coun = 0;
      } else {
        powa = arr[i];
        break;
      }
    } else {
      coun++;
    }
  }
  unsigned long long ans = 1;
  powa = min(powa, sum);
  ans = re(x, powa);
  cout << ans << endl;
  return 0;
}
