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 b[1510100], n, x;
long long Mod, C;
long long power(long long a, long long n, long long mod) {
  if (n == 0) return 1;
  long long y;
  y = power(a, n / 2, mod);
  if (n & 1) {
    return (((y * y) % mod) * a) % mod;
  } else {
    return (y * y) % mod;
  }
}
bool func(int mid) {
  int i, j, k;
  long long EX, sum;
  EX = power(x, mid, Mod);
  sum = 0;
  for (i = 0; i < n; i++) {
    if (b[i] >= mid)
      break;
    else {
      sum += power(x, b[i], EX);
      sum %= EX;
    }
  }
  if (sum == 0)
    return true;
  else
    return false;
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  scanf("%d", &x);
  long long mx, quan;
  mx = 0;
  quan = 0;
  int a[n];
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    mx = max(mx, (long long)a[i]);
    quan += a[i];
  }
  quan = quan - mx;
  for (i = 0; i < n; i++) {
    b[i] = mx - a[i];
  }
  sort(b, b + n);
  int hi, lo, mid;
  hi = mx;
  lo = 0;
  Mod = 1000000007;
  while (hi != lo) {
    mid = (hi + lo + 1) / 2;
    if (func(mid)) {
      lo = mid;
    } else {
      hi = mid - 1;
    }
  }
  quan += hi;
  cout << power(x, quan, Mod);
}
