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;
int main() {
  int n, x;
  cin >> n >> x;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  long long power_value = a[n - 1];
  int power_count = 0;
  for (int i = n - 1; i >= 0;) {
    while (i >= 0 && power_value == a[i]) power_count++, i--;
    if (power_count % x != 0) break;
    power_value--;
    power_count /= x;
  }
  while (power_count % x == 0) {
    power_value--;
    power_count /= x;
  }
  if (power_value < 0) power_value = 0;
  long long sum = 0;
  for (int i = n - 1; i >= 0; i--) sum += a[i];
  power_value = sum - power_value;
  long long gcd = 1;
  const int modulo = 1000000007;
  for (long long i = x; power_value > 0; power_value >>= 1) {
    if (power_value & 1) gcd = gcd * i % modulo;
    i = i * i % modulo;
  }
  cout << gcd;
}
