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;
int a[100100];
map<long long, int> mp;
int main() {
  int n, x;
  while (scanf("%d%d", &n, &x) != EOF) {
    int find = 0;
    long long sum = 0;
    mp.clear();
    for (int(i) = (0); (i) < (n); i++) {
      scanf("%d", &a[i]);
      sum += a[i];
    }
    for (int(i) = (0); (i) < (n); i++) {
      mp[sum - a[i]]++;
    }
    map<long long, int>::iterator it;
    while (!mp.empty()) {
      it = mp.begin();
      int k = it->second;
      int cnt = 0;
      while (k % x == 0) {
        k /= x;
        cnt++;
      }
      if (cnt == 0) {
        long long base = x;
        long long ans = 1;
        long long exp = min(sum, it->first);
        while (exp) {
          if (exp & 1) {
            ans *= base;
            if (ans >= 1000000007) ans %= 1000000007;
          }
          base *= base;
          if (base >= 1000000007) base %= 1000000007;
          exp >>= 1;
        }
        printf("%d\n", (int)ans);
        find = 1;
      }
      if (find) break;
      mp[it->first + cnt] += k;
      mp.erase(it);
    }
  }
  return 0;
}
