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>
double PI = 3.1415926535;
const int N = (int)(1 * 1e6 + 10);
using namespace std;
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % 1000000007;
    }
    b >>= 1;
    a = (a * a) % 1000000007;
  }
  return ans;
}
int main() {
  long long int n, x, i, sum = 0, mx = INT_MIN;
  cin >> n >> x;
  long long int a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
    mx = max(mx, a[i]);
  }
  map<long long int, long long int> m;
  for (i = 0; i < n; i++) {
    a[i] = mx - a[i];
    m[a[i]]++;
  }
  long long int sol = sum - mx;
  i = 0;
  while (m[i] % x == 0) {
    m[i + 1] += m[i] / x;
    i++;
  }
  sol += min(i, mx);
  cout << power(x, sol) << endl;
}
