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 a[100005];
long long powr(long long a, long long b) {
  if (!b) return 1;
  if (b == 1) return (a % mod);
  if (b % 2) return ((powr(a, b - 1) * a) % mod);
  return powr((a * a) % mod, b / 2);
}
vector<long long> v;
int main() {
  long long n, p, cnt = 0, sum = 0;
  cin >> n >> p;
  for (int i = 0; i < (int)(n); i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 0; i < (int)(n); i++) v.push_back(sum - a[i]);
  sort(v.rbegin(), v.rend());
  long long c = v[v.size() - 1];
  while (1) {
    while (v.size() > 0 && v[v.size() - 1] == c) {
      cnt++;
      v.pop_back();
    }
    if (cnt % p == 0) {
      c++;
      cnt /= p;
    } else {
      c = min(c, sum);
      cout << powr(p, c) << endl;
      return 0;
    }
  }
  return 0;
}
