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 int a[100005];
long long int mod = 1000000007;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  long long int n, m;
  cin >> n >> m;
  long long int sum = 0;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  vector<long long int> ve;
  for (long long int i = 1; i <= n; i++) {
    ve.push_back(sum - a[i]);
  }
  sort(ve.begin(), ve.end());
  long long int res = 0;
  map<long long int, long long int> ma;
  for (long long int i = 1; i <= n; i++) {
    ma[ve[i - 1]]++;
  }
  long long int curr = ve[0];
  while (1) {
    if (ma[curr] % m == 0) {
      ma[curr + 1] += (ma[curr] / m);
    } else {
      res = curr;
      break;
    }
    curr++;
  }
  res = min(res, sum);
  long long int ans = power(m, res, mod);
  cout << ans;
  return 0;
}
