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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long int a[100004];
long long int bigmod(long long int n, long long int p, long long int m) {
  if (p == 0) return 1;
  long long int x = bigmod(n, p / 2, m);
  x = (x * x) % m;
  if (p % 2 == 1) x = (x * n) % m;
  return x;
}
int main() {
  long long int x, n, i, j;
  while (cin >> n >> x) {
    long long int s = 0;
    for (i = 1; i <= n; i++) {
      cin >> a[i];
      s += a[i];
    }
    priority_queue<long long int> pq;
    for (i = 1; i <= n; i++) {
      a[i] = s - a[i];
      pq.push(-a[i]);
    }
    long long int k = 0;
    while (pq.size() > 0) {
      long long int ur = -pq.top();
      if (ur > s) break;
      long long int cnt = 0;
      while (pq.size() > 0) {
        long long int uc = -pq.top();
        if (ur == uc) {
          cnt++;
          pq.pop();
        } else
          break;
      }
      if (cnt % x == 0) {
        for (i = 0; i < (cnt / x); i++) pq.push(-ur - 1);
      } else {
        pq.push(-ur);
        break;
      }
    }
    long long int r = -pq.top();
    r = min(r, s);
    long long int ans = bigmod(x, r, 1000000007);
    cout << ans << endl;
  }
  return 0;
}
