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 MOD = 1000000007;
long long n, arr[100005], x, ans;
long long fp(long long x, long long y) {
  if (y == 0) return 1 % MOD;
  if (y % 2 == 1) return (x * fp(x, y - 1)) % MOD;
  long long r = fp(x, y / 2);
  return (r * r) % MOD;
}
int main() {
  cin >> n >> x;
  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, arr + n);
  long long cnt = 1;
  arr[n] = -10;
  for (int i = 1; i <= n; i++) {
    if (arr[i] == arr[i - 1]) {
      cnt++;
    } else {
      if (cnt % x != 0) {
        ans = min(sum, arr[i - 1]);
        break;
      } else {
        cnt /= x;
        arr[i - 1] += 1;
        i--;
      }
    }
  }
  cout << fp(x, ans);
  return 0;
}
