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 m = 1e9 + 7;
long long fastpower(long long b, long long p) {
  if (p == 0)
    return 1;
  else if (p % 2)
    return (b * fastpower((b * b) % m, p / 2) % m) % m;
  else
    return fastpower((b * b) % m, p / 2) % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  long long sum = 0, x, a;
  map<long long, long long> mop;
  cin >> n >> x;
  int arr[n + 9];
  for (int i = 0; i < n; i++) cin >> arr[i], sum += arr[i];
  for (int i = 0; i < n; i++) mop[sum - arr[i]]++;
  for (auto i : mop) {
    while (mop[i.first] >= x) mop[i.first] -= x, mop[i.first + 1]++;
  }
  for (auto i : mop) {
    if (i.second != 0) {
      cout << fastpower(x, min(i.first, sum));
      break;
    }
  }
}
