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;
using pii = pair<int, int>;
const ll mod = 1e9 + 7;
ll mod_pow(ll b, ll e) {
  ll r = 1;
  for (; e; e >>= 1, b = b * b % mod) {
    if (e & 1) r = r * b % mod;
  }
  return r;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (int &x : a) cin >> x;
  map<int, int> cnt;
  for (int x : a) cnt[-x]++;
  for (auto it = cnt.begin(); it != cnt.end();) {
    if (it->second % m) break;
    cnt[it->first + 1] += it->second / m;
    auto it2 = it;
    it++;
    cnt.erase(it2);
  }
  ll p = accumulate((a).begin(), (a).end(), 0ll);
  if (!cnt.empty() && cnt.begin()->first < 0) p += cnt.begin()->first;
  cout << mod_pow(m, p);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int tcs = 1;
  for (int tc = 1; tc <= tcs; tc++) {
    solve();
  }
}
