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 long long mod = 1000000007;
long long quick(long long x, long long mi) {
  long long res = 1;
  while (mi) {
    if (mi % 2) res = (res * x) % mod;
    x = (x * x) % mod;
    mi /= 2;
  }
  return res;
}
int main() {
  int n, x;
  int a[100010];
  scanf("%d %d", &n, &x);
  long long sum = 0;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), sum += a[i];
  sort(a, a + n);
  long long ans = 1;
  ans = (ans * quick(x, sum - a[n - 1])) % mod;
  int fenmu = a[n - 1];
  for (int i = 0; i < n; i++) a[i] = (sum - a[i]) - (sum - a[n - 1]);
  int tem = a[n - 1], now = 1;
  for (int i = n - 2; i >= 0 && fenmu; i--) {
    if (a[i] == tem)
      now++;
    else {
      int mi = 0;
      bool flag = 0;
      while (now) {
        if (now % x == 0)
          mi++, now /= x;
        else
          break;
        if (mi == a[i] - tem) {
          flag = 1;
          break;
        }
      }
      if (flag) {
        int tmp = a[i] - tem;
        while (tmp && fenmu) {
          ans = (ans * x) % mod;
          tmp--;
          fenmu--;
        }
        tem = a[i];
        now++;
      } else {
        int tmp = mi;
        while (tmp && fenmu) {
          ans = (ans * x) % mod;
          tmp--;
          fenmu--;
        }
        break;
      }
    }
  }
  while (now && fenmu) {
    if (now % x == 0) {
      if (fenmu) {
        fenmu--;
        ans = (ans * x) % mod;
      }
      now /= x;
    } else
      break;
  }
  cout << ans << endl;
  return 0;
}
