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 main() {
  long long q, w, e, r, t = 0, t1, y, i = 1000000007, a[100000];
  scanf("%I64d%I64d", &q, &y);
  for (w = 0; w < q; w++) {
    scanf("%I64d", &a[w]);
    t += a[w];
  }
  t1 = t;
  t -= a[q - 1];
  for (w = 0; w < q; w++) a[w] = a[q - 1] - a[w];
  r = e = 0;
  for (w = q - 1;;) {
    while ((w >= 0) && (a[w] - e == 0)) {
      r++;
      w--;
    }
    if (r % y == 0) {
      r /= y;
      t++;
      e++;
    } else
      break;
  }
  q = 1;
  if (t > t1) t = t1;
  for (; t; t >>= 1) {
    if (t % 2) q = (q * y) % i;
    y = (y * y) % i;
  }
  printf("%I64d", q);
  return 0;
}
