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 long long MOD = 1e9 + 7;
const int N = 1e5 + 10;
int n;
long long sum, x, a[N];
map<long long, int> cnt;
long long mpow(long long a, long long pw) {
  long long ret = 1;
  while (pw) {
    if (pw & 1) ret = (ret * a) % MOD;
    pw >>= 1;
    a = (a * a) % MOD;
  }
  return ret;
}
void add(long long v) {
  cnt[v]++;
  if (cnt[v] == x) {
    cnt.erase(v);
    add(v + 1);
  }
}
int main() {
  scanf("%d%lld", &n, &x);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    sum += a[i];
  }
  for (int i = 1; i <= n; i++) add(sum - a[i]);
  for (auto &it : cnt) sum = min(sum, it.first);
  printf("%lld", mpow(x, sum));
  return 0;
}
