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;
using LL = long long;
const int N = 1e5 + 4, M = 1e9 + 7;
int n, x, a[N];
int fast_pow(int a, LL e) {
  int b = 1;
  for (LL i = e; i; i >>= 1, a = (LL)a * a % M) {
    if (i & 1) b = (LL)b * a % M;
  }
  return b;
}
int main() {
  scanf("%d%d", &n, &x);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  reverse(a + 1, a + n + 1);
  LL sum = 0;
  for (int i = 1; i <= n; i++) sum += a[i];
  int i = 1, carry = 0, v = a[1];
  while (i <= n && v > 0) {
    if (v > a[i]) {
      if (carry) {
        if (carry % x != 0) {
          carry = 0;
          break;
        } else {
          carry /= x;
          v--;
        }
      } else
        v = a[i];
    } else {
      int j = i;
      for (; j <= n && a[j] == a[i]; j++) carry++;
      if (carry % x != 0) {
        carry = 0;
        break;
      } else
        carry /= x;
      v = a[i] - 1;
      i = j;
    }
  }
  while (carry && v > 0) {
    if (carry % x != 0)
      break;
    else {
      v--;
      carry /= x;
    }
  }
  printf("%d\n", fast_pow(x, (sum % (M - 1) - (v + M - 1) % (M - 1) + M - 1) %
                                 (M - 1)));
}
