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 MOD = 1e9 + 7;
long long pow(long long x, long long v) {
  if (v == 0) return 1LL;
  if (v % 2) return (x % MOD) * pow(x % MOD, v - 1) % MOD;
  return pow((x * x) % MOD, v / 2) % MOD;
}
int main() {
  int n;
  long long x, sum = 0, v, mx = 0;
  scanf("%d%lld", &n, &x);
  map<long long, int> cnt;
  vector<long long> a(n);
  for (int i = 0; i < n; ++i)
    scanf("%lld", &a[i]), sum += a[i], mx = max(mx, a[i]);
  for (int i = 0; i < n; ++i) cnt[sum - a[i]]++;
  for (map<long long, int>::iterator it = cnt.begin(); it != cnt.end(); ++it) {
    v = it->first;
    if (it->second % x == 0) {
      cnt[v + 1] += it->second / x;
    } else
      break;
  }
  v = min(v, sum);
  printf("%lld", pow(x, v));
  return 0;
}
