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 a[100000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x;
  cin >> n >> x;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long sum = 0;
  for (int i = 0; i < n; i++) sum += a[i];
  for (int i = 0; i < n; i++) a[i] = sum - a[i];
  sort(a, a + n);
  stack<pair<long long, long long> > s;
  for (int i = n - 1; i >= 0; i--) s.push(make_pair(a[i], 1));
  long long ans;
  long long k;
  while (!s.empty() && ans < sum) {
    ans = (s.top()).first;
    k = 0;
    while (!s.empty() && (s.top()).first == ans) {
      k += (s.top()).second;
      s.pop();
    }
    if (k % x == 0) {
      s.push(make_pair(ans + 1, k / x));
    } else
      break;
  }
  long long mul = x;
  long long ans1 = 1;
  while (ans != 0) {
    if (ans % 2 == 1) ans1 = (ans1 * mul) % (long long)(1e9 + 7);
    mul = (mul * mul) % (long long)(1e9 + 7);
    ans /= 2;
  }
  cout << ans1 << endl;
  return 0;
}
