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 a[100010], sum, sum1, b, cnt;
multiset<long long> s;
int binpow(long long a, long long n) {
  long long res = 1;
  while (n)
    if (n & 1) {
      res *= a;
      res %= 1000000007;
      --n;
    } else {
      a *= a;
      a %= 1000000007;
      n >>= 1;
    }
  return res;
}
int main() {
  int n, x;
  cin >> n >> x;
  for (int i = 1; i <= n; i++) cin >> a[i], sum += a[i];
  reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) s.insert(a[1] - a[i]);
  sum1 = sum - a[1];
  while (true) {
    b = *s.begin();
    cnt = s.size();
    s.erase(b);
    cnt = cnt - s.size();
    if (cnt % x == 0) {
      for (int i = 1; i <= cnt / x; i++) s.insert(b + 1);
    } else {
      sum1 += b;
      break;
    }
  }
  cout << binpow(x, min(sum1, sum));
  return 0;
}
