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 N = 100 * 100 * 10 + 11;
const int mod = 1000 * 1000 * 1000 + 7;
long long a[N], s;
long long n, m, p;
long long pow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * x) % mod;
    x = x * x % mod;
    y /= 2;
  }
  return ret;
}
int main() {
  cin >> n >> p;
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 1; i <= n; i++) a[i] = sum - a[i];
  long long mn = sum;
  long long now = a[n];
  long long tot = 0;
  while (n) {
    if (a[n] == now) {
      tot++;
      n--;
    } else {
      if (tot % p) break;
      tot /= p;
      now++;
    }
  }
  while (tot && (tot % p == 0)) {
    now++;
    tot /= p;
  }
  cout << pow(p, min(mn, now)) % mod << endl;
  return 0;
}
