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 n, x;
long long cal(long long a) {
  if (a == 0) return 1;
  if (a % 2) return (x * cal(a - 1)) % 1000000007;
  long long ans = cal(a / 2);
  return (ans * ans) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x;
  vector<long long> arr(n);
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  for (int i = 0; i < n; i++) {
    arr[i] = sum - arr[i];
  }
  sort(arr.rbegin(), arr.rend());
  while (true) {
    long long ans = arr[arr.size() - 1];
    int cnt = 0;
    while (arr.size() > 0 && arr[arr.size() - 1] == ans) {
      cnt++;
      arr.pop_back();
    }
    if (cnt % x != 0) {
      cout << cal(min(ans, sum)) << endl;
      return 0;
    } else {
      for (int i = 0; i < cnt / x; i++) {
        arr.push_back(ans + 1);
      }
    }
  }
  return 0;
}
