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 A[100008];
map<long long, long long> mymap;
long long pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return res;
}
int main() {
  long long n, x;
  long long ans, total = 0;
  scanf("%lld", &n);
  scanf("%lld", &x);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &A[i]);
    total += A[i];
  }
  for (int i = 0; i < n; i++) {
    A[i] = total - A[i];
    mymap[A[i]]++;
  }
  for (typeof((mymap).begin()) it = (mymap).begin(); it != (mymap).end();
       it++) {
    if (it->second / x != 0) {
      mymap[it->first + 1] += it->second / x;
    }
    mymap[it->first] = it->second % x;
    if (mymap[it->first] != 0) {
      ans = it->first;
      break;
    }
  }
  ans = pow(x, min(ans, total));
  printf("%lld\n", ans);
  ;
  return 0;
}
