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 powmod(long long a, long long b, long long MOD) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
const int N = 1e5 + 5;
int a[N];
map<long long, int> m;
long long getAns(long long n) {
  long long x = m.begin()->first;
  while (m[x] % n == 0) {
    m[x + 1] += m[x] / n;
    x++;
  }
  return x;
}
int main() {
  int n;
  long long x, sum = 0;
  scanf("%d%I64d", &n, &x);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    sum += a[i];
  }
  for (int i = 0; i < n; i++) {
    m[sum - a[i]]++;
  }
  sum = min(getAns(x), sum);
  printf("%I64d\n", powmod(x, sum, 1e9 + 7));
  return 0;
}
