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;
constexpr int mod = 1e9 + 7;
ll modpow(ll x, ll n) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) (res *= x) %= mod;
    (x *= x) %= mod;
    n >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ll n, x;
  cin >> n >> x;
  vector<ll> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  const ll an = a.back();
  ll ans = modpow(x, accumulate(begin(a), end(a) - 1, 0LL));
  ll m = 0;
  map<ll, ll> cnt;
  for (int i = 0; i < n; ++i) {
    cnt[an - a[i]] += 1;
  }
  int last = 0;
  while (true) {
    auto p = *cnt.begin();
    cnt.erase(begin(cnt));
    if (p.second % x == 0) {
      cnt[p.first + 1] += p.second / x;
      last = p.first + 1;
    } else {
      m = min(an, m + last);
      break;
    }
  }
  ans = (ans * modpow(x, m)) % mod;
  cout << ans << endl;
}
