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;
unsigned long long fp(unsigned long long a, unsigned long long b) {
  if (!b) return 1;
  unsigned long long u = fp(a, b / 2);
  unsigned long long r = ((u % 1000000007) * (u % 1000000007)) % 1000000007;
  if (b % 2 == 0) return r;
  return (r * a % 1000000007) % 1000000007;
}
map<long long, long long> freq;
long long start = 1e18, sum;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x, cnt = 0, sub = 0;
  cin >> n >> x;
  long long arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i], sum += arr[i];
  for (int i = 0; i < n; i++)
    freq[sum - arr[i]]++, (start = min(start, sum - arr[i]));
  cnt += start;
  if (freq[start] % x != 0) goto go;
  sub += start;
  freq[1] = freq[start] / x + freq[sub + 1];
  while (1) {
    cnt++;
    sub++;
    if (freq[1] % x != 0) break;
    freq[1] = freq[1] / x + freq[sub + 1];
  }
go:;
  cout << fp(x, min(sum, cnt));
  return 0;
}
