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 N = 2e5;
long long bp(long long a, long long b) {
  if (b == 0) return 1;
  long long c = bp(a, b / 2);
  c = (c * c) % 1000000007;
  if (b % 2) c = (c * a) % 1000000007;
  return c;
}
int n;
long long X, sum, a[N];
multiset<long long> second;
int main() {
  scanf("%d%I64d", &n, &X);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    sum += a[i];
  }
  for (int i = 1; i <= n; ++i) second.insert(sum - a[i]);
  long long ans = sum;
  while (!second.empty()) {
    int k = 1;
    long long x = *second.begin();
    second.erase(second.find(x));
    while (!second.empty() && *second.begin() == x) {
      k++;
      second.erase(second.find(x));
    }
    if (k % X != 0) {
      ans = min(ans, (long long)x);
      break;
    }
    k /= X;
    for (int i = 0; i < k; i++) second.insert(x + 1);
  }
  ans = bp(X, ans);
  printf("%I64d", ans);
  return 0;
}
