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 power(long long a, long long p) {
  long long ret = 1;
  a %= 1000000007;
  while (p) {
    if (p % 2 == 1) {
      ret = ret * a;
      ret %= 1000000007;
    }
    a = a * a;
    a %= 1000000007;
    p /= 2;
  }
  return ret;
}
int main(void) {
  int n, x;
  cin >> n >> x;
  vector<long long> vec(n);
  long long s = 0;
  for (int i = 0; i < n; i++) {
    cin >> vec[i];
    s += vec[i];
  }
  for (int i = 0; i < n; i++) vec[i] = s - vec[i];
  sort(vec.rbegin(), vec.rend());
  while (true) {
    int ind = vec.size() - 1;
    long long num = vec[ind];
    int c = 0;
    while (vec[ind - c] == num && ind - c >= 0) {
      vec.pop_back();
      c++;
    }
    if (c % x != 0) {
      long long p = min(num, s);
      cout << power(x, p);
      return 0;
    }
    for (int i = 0; i < c / x; i++) vec.push_back(num + 1);
  }
}
