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;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 9;
long long poww(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * x % mod;
    n >>= 1;
    x = x * x % mod;
  }
  return ans;
}
long long a[maxn];
int main() {
  ios::sync_with_stdio(false);
  int n;
  int x;
  while (cin >> n >> x) {
    long long sum = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    sort(a + 1, a + 1 + n);
    long long ans = sum - a[n];
    long long cnt = 0;
    for (int i = a[n]; i >= 1; i--) {
      while (n && a[n] == i) {
        n--;
        cnt++;
      }
      if (cnt % x) break;
      ans++;
      cnt /= x;
    }
    cout << poww(x, ans) << endl;
  }
  return 0;
}
