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[121212];
priority_queue<int> Q;
long long mod = 1e9 + 7;
long long po(long long x, long long y) {
  int i;
  long long res = 1;
  for (i = 0; i < 62; i++) {
    if (((long long)1 << i) & y) res = (res * x) % mod;
    x = (x * x) % mod;
  }
  return res;
}
int main() {
  int n, x;
  int i;
  long long S = 0;
  scanf("%d%d", &n, &x);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    S += a[i];
    Q.push(a[i]);
  }
  int cnt;
  int now;
  while (!Q.empty()) {
    cnt = 0;
    now = Q.top();
    if (now == 0) break;
    for (i = 0; i < x; i++) {
      if (Q.empty()) break;
      int tt = Q.top();
      Q.pop();
      if (now == tt)
        cnt++;
      else
        break;
    }
    if (cnt != x) break;
    Q.push(now - 1);
  }
  printf("%lld", po(x, S - now));
  return 0;
}
