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>
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
std::pair<int, int> operator+(std::pair<int, int> a, std::pair<int, int> b) {
  return {a.first + b.first, a.second + b.second};
}
std::pair<long long, long long> operator+(std::pair<long long, long long> a,
                                          std::pair<long long, long long> b) {
  return {a.first + b.first, a.second + b.second};
}
std::ostream& operator<<(std::ostream& out, std::pair<int, int> a) {
  out << a.first << " " << a.second << "\n";
  return out;
}
std::ostream& operator<<(std::ostream& out, std::pair<long long, long long> a) {
  out << a.first << " " << a.second << "\n";
  return out;
}
std::istream& operator>>(std::istream& in, std::pair<int, int>& a) {
  in >> a.first >> a.second;
  return in;
}
std::istream& operator>>(std::istream& in, std::pair<long long, long long>& a) {
  in >> a.first >> a.second;
  return in;
}
using namespace std;
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res % 1000000007;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, x;
  cin >> n >> x;
  int a[n];
  for (int i = 0; i < n; ++i) cin >> a[i];
  map<long long, int> cnt;
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += a[i];
  }
  for (int i = 0; i < n; ++i) {
    cnt[sum - a[i]]++;
  }
  for (auto& b : cnt) {
    if (b.second % x == 0) {
      cnt[b.first + 1] += b.second / x;
      cnt[b.first] = 0;
    }
  }
  for (auto& b : cnt) {
    if (b.second != 0) {
      cout << binpow(x, min(b.first, sum)) << "\n";
      return 0;
    }
  }
}
