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;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (T i : v) os << i << ", ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (T i : v) os << i << " ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  for (T i : v) os << i << " ";
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const pair<T, second> &v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const map<T, second> &v) {
  for (auto i : v) os << '(' << i.first << " => " << i.second << ')' << ' ';
  return os;
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const long long MOD = 1e9 + 7;
vector<long long> v;
long long n, k;
long long power(long long x, long long y, long long p = MOD) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long sum = 0;
  for (long long i : v) sum += i;
  vector<long long> temp;
  for (long long i : v) temp.push_back(sum - i);
  long long p = temp.back();
  while (1) {
    long long t = temp.back();
    long long c = 0;
    while (!temp.empty() && temp.back() == t) {
      c++;
      temp.pop_back();
    }
    if (c % k == 0 && c > 0) {
      c /= k;
      while (c--) temp.push_back(t + 1);
    } else {
      p = t;
      break;
    }
  }
  cout << power(k, min(p, sum)) << "\n";
}
int main() {
  fast();
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long a;
    cin >> a;
    v.push_back(a);
  }
  solve();
}
