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 pow(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
int main() {
  int n, equal = 1;
  long long x, y[100002], sum = 0, MOD = 1000000007;
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> y[i];
    if (y[i] == y[i - 1]) {
      equal++;
    } else {
      equal = 1;
    }
    sum += y[i];
  }
  if (y[n - 1] == 0)
    cout << "1";
  else {
    map<int, int> m;
    m.clear();
    long long last = y[n - 1];
    long long ans, xy;
    for (int i = n - 1; i >= 0; i--) {
      xy = last - y[i];
      m[xy]++;
      while (m[xy] == x) {
        m[xy] = 0;
        m[xy + 1]++;
        xy++;
      }
    }
    for (int i = 0;; i++)
      if (m[i]) {
        ans = i;
        break;
      }
    ans = min(ans, last);
    cout << pow(x, sum - last + ans, MOD);
  }
  return 0;
}
