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 pw(long long x, long long n) {
  if (n == 0) return 1;
  if (n % 2)
    return x * pw(x, n - 1) % 1000000007;
  else {
    long long ret = pw(x, n / 2);
    return ret * ret % 1000000007;
  }
}
map<long long, long long> d;
int n;
long long x, a[100001];
int idx;
int main() {
  scanf("%d %lld", &n, &x);
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    sum += a[i];
  }
  for (int i = 0; i < n; i++) {
    d[sum - a[i]]++;
  }
  while (1) {
    map<long long, long long>::iterator it = d.begin();
    if (it->second % x) {
      printf("%lld\n", pw(x, min(sum, it->first)));
      return 0;
    }
    d[it->first + 1] += it->second / x;
    d.erase(it);
  }
}
