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 int nm = 100002;
const long long mod = 1000000007;
int n;
long long x;
long long a[nm];
void nhap() {
  scanf("%d%I64d", &n, &x);
  int i;
  for (i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
}
long long power(long long a, long long n) {
  if (!n) return 1;
  long long tg = power(a, n / 2);
  tg = (tg * tg) % mod;
  if (n % 2) tg = (tg * a) % mod;
  return tg;
}
void xuli() {
  long long sum = 0;
  int i;
  for (i = 1; i <= n; ++i) sum += a[i];
  for (i = 1; i <= n; ++i) a[i] = sum - a[i];
  sort(a + 1, a + n + 1);
  int j, k, u;
  i = 1;
  do {
    j = i;
    while (j + 1 <= n && a[j + 1] == a[j]) ++j;
    if ((j - i + 1) % x == 0) {
      k = (j - i + 1) / x;
      for (u = 1; u <= k; ++u) a[j + 1 - u] = a[i] + 1;
      i = j + 1 - k;
    } else {
      printf("%I64d", power(x, min(a[i], sum)));
      return;
    }
  } while (1);
}
int main() {
  nhap();
  xuli();
}
