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 M = 1e3 + 5;
long long fp(long long base, long long exp) {
  if (exp == 0) return 1;
  long long ans = fp(base, exp / 2);
  ans = (ans * ans) % 1000000007;
  if (exp % 2 != 0) ans = (ans * (base % 1000000007)) % 1000000007;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  long long x, sum = 0, arr[n + 5], cyc;
  cin >> x;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    mp[sum - arr[i]]++;
  }
  for (auto i : mp) {
    cyc = i.second / x;
    if (i.second % x) {
      cout << fp(x, min(sum, i.first)) % 1000000007;
      break;
    }
    mp[i.first + 1] += cyc;
    i.second %= x;
  }
  return 0;
}
