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 long long N = 100005;
const long long MOD = 1000000007;
long long n, x;
long long a[N];
map<long long, long long> w;
long long quick_mod(long long a, long long b, long long m) {
  long long ans = 1;
  a %= m;
  while (b) {
    if (b & 1) {
      ans = ans * a % m;
      b--;
    }
    b >>= 1;
    a = a * a % m;
  }
  return ans;
}
int main() {
  long long s = 0;
  w.clear();
  scanf("%I64d%I64d", &n, &x);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    s += a[i];
    w[a[i]]++;
  }
  int i = a[n - 1];
  while (w[i] % x == 0) {
    w[i - 1] += w[i] / x;
    --i;
  }
  if (i < 0) i = 0;
  printf("%I64d\n", quick_mod(x, s - i, MOD));
  return 0;
}
