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;
unsigned long long int ex(unsigned long long int base,
                          unsigned long long int po) {
  if (base == 0) return 0;
  if (po == 0) return 1;
  unsigned long long int ans = ex(base, po / 2) % 1000000007;
  if (po & 1)
    return ans * ans % 1000000007 * base % 1000000007;
  else
    return ans * ans % 1000000007;
}
map<unsigned long long int, unsigned long long int> m;
map<unsigned long long int, unsigned long long int>::iterator it;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  int n, x, i;
  unsigned long long int sum = 0, cur;
  cin >> n >> x;
  vector<unsigned long long int> po;
  vector<unsigned long long int> div;
  div.push_back(1);
  while (div.back() < (unsigned long long int)1e18)
    div.push_back(div.back() * x);
  for (i = 0; i < n; i++) {
    cin >> cur;
    sum += cur, po.push_back(cur);
  }
  for (i = 0; i < n; i++) {
    cur = sum - po[i];
    if (m.find(cur) == m.end())
      m[cur] = 1;
    else
      m[cur]++;
  }
  for (it = m.begin(); it != m.end(); it++) {
    cur = it->first;
    unsigned long long int val = it->second;
    int ind = 0;
    while (val % div[ind] == 0) {
      ind++;
    }
    if (--ind) {
      if (m.find(cur + ind) == m.end())
        m.insert({cur + ind, val / div[ind]});
      else
        m[cur + ind] += val / div[ind];
    } else
      break;
  }
  cout << ex(x, min(cur, sum)) << endl;
  return 0;
}
