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 n, x, num = 1e18, den, mod = 1e9 + 7, a[100005];
map<long long, long long> mp;
long long fp(long long b, long long p, long long mod) {
  if (!p) return 1;
  long long res = fp(b, p / 2, mod);
  res = (res % mod * res % mod) % mod;
  if (p & 1) res = (res % mod * b % mod) % mod;
  return res % mod;
}
int main() {
  scanf("%lld%lld", &n, &x);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]), den += a[i];
  for (int i = 0; i < n; i++) mp[den - a[i]]++;
  for (auto i : mp) {
    long long f = i.first, s = i.second, cnt = 0;
    cnt = s / x;
    if (cnt) mp[f + 1] += cnt;
    s -= x * cnt;
    if (s > 0) num = min(num, f);
  }
  printf("%lld\n", fp(x, min(num, den), mod));
  return 0;
}
