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;
map<long long int, long long int> cnt;
map<long long int, long long int>::iterator it;
long long int a[1000001];
long long int bpow(long long int x, long long int n) {
  long long int ans = 1;
  while (n > 0) {
    if (n & 1) ans *= x;
    x *= x;
    ans %= 1000000007;
    x %= 1000000007;
    n /= 2;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  ;
  cin.tie(0);
  ;
  long long int i, j, n, x, sum = 0, ctr, power, ans;
  cin >> n >> x;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  ans = sum;
  for (i = 1; i <= n; i++) cnt[sum - a[i]]++;
  while (1) {
    it = cnt.begin();
    ctr = it->second;
    power = it->first;
    if (ctr % x == 0) {
      cnt[power + 1] += ctr / x;
      cnt.erase(it);
    } else {
      ans = min(ans, power);
      break;
    }
  }
  cout << bpow(x, ans);
  return 0;
}
