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;
long long a[110000], s, b[110000];
int c[110000];
int n, x;
const int mod = 1000000007;
int mul(int a, long long b) {
  int res = 1;
  for (; b; b /= 2, a = a * (long long)a % mod)
    if (b & 1) res = res * (long long)a % mod;
  return res;
}
int main() {
  scanf("%d%d", &n, &x);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]), s += a[i];
  for (int i = 1; i <= n; ++i) a[i] = s - a[i];
  sort(a + 1, a + 1 + n);
  b[1] = a[1], c[1] = 1;
  int t = 1;
  for (int i = 2; i <= n; ++i)
    if (a[i] == b[t])
      ++c[t];
    else
      b[++t] = a[i], c[t] = 1;
  long long res = a[1];
  for (int i = 1; i <= t; ++i)
    if (c[i] % x != 0) {
      res = b[i];
      break;
    } else if (b[i] != b[i + 1] - 1)
      ++b[i], c[i--] /= x;
    else
      c[i + 1] += c[i] / x;
  printf("%d", mul(x, min(res, s)));
}
