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 res;
int x;
int n, mas[100002];
long long sum;
map<long long, long long> m;
long long power(long long a, long long b) {
  long long res = 1;
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long c = power(a, b / 2) % 1000000007;
    return (c * c) % 1000000007;
  } else
    return (power(a, b - 1) * a) % 1000000007;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> n >> x;
  for (int i = 0; i < n; ++i) {
    cin >> mas[i];
    sum += mas[i];
  }
  for (int i = 0; i < n; ++i) m[sum - mas[i]]++;
  long long countt = 1, minn = 1000000000009;
  int j = 0;
  for (map<long long, long long>::iterator it = m.begin(); it != m.end();
       ++it) {
    if ((*it).second % x != 0) {
      cout << power(x, min(sum, (*it).first));
      return 0;
    }
    m[(*it).first + 1] += (*it).second / x;
    minn = (*it).first + 1;
  }
  cout << power(x, min(minn, sum));
  return 0;
}
