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 power(long long a, long long n) {
  long long s = 1;
  while (n) {
    if (n & 1) s = s * a % 1000000007;
    a = a * a % 1000000007;
    n >>= 1;
  }
  return s;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long c, i, j, k, l, m, t, x, y, n, z;
  string s;
  cin >> n >> x;
  long long a1[n];
  priority_queue<long long, vector<long long>, greater<long long> > pq;
  for (i = 0; i < n; i++) cin >> a1[i];
  long long sum = 0;
  for (i = 0; i < n; i++) sum = sum + a1[i];
  z = 0;
  map<long long, long long> mp;
  for (i = 0; i < n; i++) mp[sum - a1[i]]++;
  while (mp.begin()->second % x == 0) {
    mp[mp.begin()->first + 1] += mp.begin()->second / x;
    mp.erase(mp.begin());
  }
  long long b = mp.begin()->first;
  cout << power(x, min(sum, b)) << endl;
  return 0;
}
