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;
long long a[100005], x, tot = 0, ans, mod = 1e9 + 7, res;
vector<pair<long long, long long>> rev;
stack<pair<long long, long long>> waiting;
int n;
void createRev() {
  rev.push_back({-1, -1});
  int id = 0;
  for (int i = n; i > 0; i--) {
    long long ins = tot - a[i];
    if (rev[id].first == ins) {
      rev[id].second++;
    } else {
      rev.push_back({ins, 1});
      id++;
    }
  }
}
void findtop() {
  for (int i = 1; i < rev.size(); i++) {
    if (!waiting.empty()) {
      while (waiting.top().first < rev[i].first) {
        if (waiting.top().second % x != 0) {
          ans = waiting.top().first;
          return;
        }
        pair<long long, long long> temp = waiting.top();
        waiting.pop();
        waiting.push({temp.first + 1, temp.second / x});
      }
      if (waiting.top().first == rev[i].first) {
        rev[i].second += waiting.top().second;
        waiting.pop();
      }
    }
    if (rev[i].second % x != 0) {
      ans = rev[i].first;
      return;
    } else {
      waiting.push({rev[i].first + 1, rev[i].second / x});
    }
  }
  if (!waiting.empty()) {
    while (waiting.top().second % x == 0) {
      pair<long long, long long> temp = waiting.top();
      waiting.pop();
      waiting.push({temp.first + 1, temp.second / x});
    }
    ans = waiting.top().first;
    return;
  }
  return;
}
long long pngkt(long long p) {
  long long temp;
  if (p == 0)
    return 1;
  else if (p % 2 == 1) {
    temp = pngkt(p / 2) % mod;
    return (((temp * temp) % mod) * x) % mod;
  } else {
    temp = pngkt(p / 2) % mod;
    return ((temp * temp) % mod);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> x;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    tot += a[i];
  }
  sort(a + 1, a + 1 + n);
  createRev();
  findtop();
  ans = min(ans, tot);
  res = pngkt(ans);
  cout << res << endl;
  return 0;
}
