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 base = 1e9;
const int maxn = 1e3;
const int logn = 20;
const int inf = 2e9;
const int alpha = 26;
const int my_hash = 29;
const long long l_inf = 4e18;
const long long mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-9;
int n;
long long x, sum = 0;
vector<long long> a, b;
long long binpow(long long a, long long b) {
  if (!b) return 1;
  if (b % 2) return (binpow(a, b - 1) * a) % mod;
  long long k = binpow(a, b / 2);
  return (k * k) % mod;
}
int main() {
  ios_base::sync_with_stdio();
  cin >> n >> x;
  a = vector<long long>(n);
  for (int i = int(0); i < int(n); i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = int(0); i < int(n); i++) b.push_back(sum - a[i]);
  sort(b.rbegin(), b.rend());
  while (true) {
    long long cnt = 0, v = b.back();
    while (!b.empty() && b.back() == v) {
      cnt++;
      b.pop_back();
    }
    if (cnt % x) {
      cout << binpow(x, min(v, sum));
      exit(0);
    } else {
      cnt /= x;
      for (int i = int(0); i < int(cnt); i++) b.push_back(v + 1);
    }
  }
}
