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;
const int maxn = 100000 + 10;
const int mod = 1000000000 + 7;
long long num[maxn], a[maxn];
long long f(long long x, long long y) {
  long long res = 1, b = x;
  while (y) {
    if (y & 1) res = (res * b) % mod;
    b = (b * b) % mod;
    y >>= 1;
  }
  return res;
}
long long cal(long long x, long long y) {
  long long ret = 0;
  while (x && (x % y == 0)) {
    ret++;
    x /= y;
  }
  return ret;
}
int main() {
  int n, x;
  while (~scanf("%d%d", &n, &x)) {
    long long sum = 0;
    for (int i = 0; i < n; ++i) {
      scanf("%I64d", &a[i]);
      sum += a[i];
    }
    for (int i = 0; i < n; ++i) num[i] = sum - a[i];
    sort(num, num + n);
    long long cnt = 0, ont = 1, op = num[0], tmp;
    int p = 0;
    while (true) {
      while (p + 1 < n && num[p + 1] - cnt == op) ont++, p++;
      cnt += op;
      if ((ont % x) || (op == 0)) {
        tmp = min(cnt, sum);
        long long ans = f(x, tmp);
        printf("%I64d\n", ans);
        break;
      }
      op = cal(ont, x);
      if (p + 1 < n) op = min(op, num[p + 1] - cnt);
      for (int i = 0; i < op; ++i) ont /= x;
    }
  }
  return 0;
}
