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 mod = 1e9 + 7;
const int N = 1e5 + 10;
long long a[N], pw[N];
long long poww(long long x, long long y) {
  if (y == 0) return 1ll;
  long long ans = poww(x, y / 2);
  if (y % 2 == 0) return ans * ans % mod;
  return ans * ans % mod * x % mod;
}
int main() {
  long long n, x;
  cin >> n >> x;
  long long sum = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[n + 1 - i];
    sum += a[n + 1 - i];
  }
  for (long long i = 1; i <= n; i++) {
    a[i] = sum - a[i];
  }
  long long t = 0;
  pw[0] = 1;
  long long i = 1;
  while (true) {
    pw[i] = pw[i - 1] * x;
    i++;
    if (pw[i - 1] >= 1e9) break;
  }
  long long mx_pw = i - 1;
  long long l = 0, r = sum + 1;
  while (r - l > 1) {
    long long mid = (l + r) / 2;
    bool flag = true;
    long long t = 0;
    long long last = a[1];
    long long indx = n;
    for (long long i = 1; i <= n; i++) {
      if (a[i] >= mid) {
        indx = i - 1;
        break;
      }
      if (a[i] == last) {
        t++;
        continue;
      }
      long long tavan = min(a[i] - last, mx_pw);
      if (t % pw[tavan] != 0) {
        flag = false;
        break;
      }
      t /= pw[tavan];
      t++;
      last = a[i];
    }
    if (indx != 0) {
      if (t % pw[min(mx_pw, mid - a[indx])] != 0) {
        flag = false;
      }
    }
    if (flag)
      l = mid;
    else
      r = mid;
  }
  cout << poww(x, l);
}
