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 fastExpo(long long x, long long n) {
  long long temp = 1;
  while (n != 0) {
    if (n & 1) temp = (temp * x) % 1000000007;
    x *= x;
    x %= 1000000007;
    n /= 2;
  }
  return (temp % 1000000007);
}
int main() {
  long long n, i, j, x;
  cin >> n >> x;
  long long arr[n];
  long long sum = 0;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  vector<long long> siir;
  for (i = 0; i < n; i++) siir.push_back(sum - arr[i]);
  sort(siir.begin(), siir.end());
  reverse(siir.begin(), siir.end());
  while (1) {
    long long y = siir[siir.size() - 1];
    int cnt = 0;
    while (siir.size() > 0 && siir[siir.size() - 1] == y) {
      cnt++;
      siir.pop_back();
    }
    if (cnt % x) {
      y = min(y, sum);
      cout << fastExpo(x, y % (1000000007 - 1)) % 1000000007 << endl;
      return 0;
    } else {
      cnt /= x;
      for (i = 0; i < cnt; i++) siir.push_back(y + 1);
    }
  }
}
