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;
void debug_out() { cerr << endl; }
template <class T>
ostream& prnt(ostream& out, T v) {
  out << v.size() << '\n';
  for (auto e : v) out << e << ' ';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> v) {
  return prnt(out, v);
}
template <class T>
ostream& operator<<(ostream& out, set<T> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ' ' << p.second << ')';
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
const long long N = 100100;
const long long MOD = 1e9 + 7;
long long n, a[N], s, k;
map<long long, long long> m;
long long powMod(long long base, long long exp) {
  if (exp == 0) return 1;
  long long tmp = powMod(base, exp / 2);
  tmp = (1LL * tmp * tmp) % MOD;
  if (exp % 2) tmp = (tmp * base) % MOD;
  return tmp;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i], s += a[i];
  for (int i = 1; i <= n; i++) m[s - a[i]]++;
  while (m.begin()->second % k == 0) {
    pair<long long, long long> p = *m.begin();
    p.first++;
    p.second /= k;
    m.erase(m.begin());
    m[p.first] += p.second;
  }
  long long off = min(m.begin()->first, s);
  cout << powMod(k, off);
}
