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;
using ll = long long;
const ll MOD = 1000000007LL;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int n, x;
  cin >> n >> x;
  vector<int> arr(n);
  ll sum = 0LL;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    sum += 1LL * arr[i];
  }
  map<ll, int> cnt;
  for (int i = 0; i < n; ++i) {
    ++cnt[sum - arr[i]];
  }
  ll val;
  auto it = cnt.begin();
  while (true) {
    if (it->second % x) {
      val = it->first;
      break;
    }
    cnt[it->first + 1] += (it->second / x);
    ++it;
  }
  val = min(val, sum);
  vector<ll> lg2(64);
  lg2[0] = x;
  for (int i = 1; i < 64; ++i) {
    lg2[i] = (lg2[i - 1] * lg2[i - 1]) % MOD;
  }
  ll sol = 1LL;
  while (val) {
    ll lg = ll(log2(val));
    val -= (1LL << lg);
    sol = (sol * lg2[lg]) % MOD;
  }
  cout << sol;
  return 0;
}
