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;
map<int, int> tree;
map<int, int>::iterator iter;
int n, x, a;
long long sum = 0;
long long mod = 1000000007;
long long po(long long a, long long s) {
  if (s == 0) return 1;
  if (s & 1)
    return po(a * a % mod, s >> 1) * a % mod;
  else
    return po(a * a % mod, s >> 1);
}
void in(int s) {
  iter = tree.find(s);
  if (iter == tree.end()) {
    tree.insert(make_pair(s, 1));
  } else {
    iter->second++;
    if (iter->second == x) {
      tree.erase(iter);
      in(s + 1);
    }
  }
}
int main() {
  tree.clear();
  scanf("%d%d", &n, &x);
  for (int q = 0; q < n; q++) {
    scanf("%d", &a);
    sum += a;
    in(-a);
  }
  iter = tree.begin();
  cout << po(x, min(sum + iter->first, sum)) << endl;
  return 0;
}
