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 binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b = b >> 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(10);
  long long int t = 1;
  while (t--) {
    long long int n, x;
    cin >> n >> x;
    long long int a[n], sum = 0;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    vector<long long int> sumu;
    for (long long int i = 0; i < n; i++) {
      sumu.push_back(sum - a[i]);
    }
    sort(sumu.rbegin(), sumu.rend());
    while (1) {
      long long int y = sumu[sumu.size() - 1];
      long long int count = 0;
      while (sumu.size() > 0 && y == sumu[sumu.size() - 1]) {
        count++;
        sumu.pop_back();
      }
      if (count % x != 0) {
        y = min(sum, y);
        cout << binpow(x, y);
        return 0;
      } else {
        count /= x;
        while (count--) sumu.push_back(y + 1);
      }
    }
  }
  return 0;
}
