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 long long MOD = 1000000007;
long long pw(long long a, long long b, long long m = MOD) {
  if (b == 0) return 1;
  if (b & 1) return a * pw(a, b - 1) % m;
  return pw(a * a % m, b / 2);
}
void solve() {
  long long n, x;
  cin >> n >> x;
  vector<long long> v(n);
  long long sum = 0;
  for (long long i = 0; i < n; i++) cin >> v[i], sum += v[i];
  long long ans = 0;
  long long tp = *max_element(v.begin(), v.end());
  map<long long, long long> cnt;
  for (long long i = 0; i < n; i++) {
    v[i] = tp - v[i];
    cnt[v[i]]++;
  }
  ans += (sum - tp);
  long long i = 0;
  while (cnt[i] % x == 0) {
    cnt[i + 1] += cnt[i] / x;
    i++;
  }
  ans = pw(x, ans + min(i, tp));
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t = 1;
  while (t--) solve();
  return 0;
}
