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;
const int MOD = 1e9 + 7;
const long long INF = 1e18;
const int MAXN = 1e5 + 5;
const int T = 1e3 + 10;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
long long binpow(long long val, long long deg) {
  if (!deg) return 1;
  if (deg & 1) return binpow(val, deg - 1) * val % MOD;
  long long res = binpow(val, deg / 2);
  return (res * res) % MOD;
}
long long inv(long long a) { return binpow(a, MOD - 2); }
pair<int, int> operator+(const pair<int, int>& a, const pair<int, int>& b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<int, int> operator-(const pair<int, int>& a, const pair<int, int>& b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
void rotate90(int& first, int& second, int x0, int y0) {
  first = first - x0;
  second = second - y0;
  swap(first, second);
  first = -first;
  first += x0;
  second += y0;
}
long long i, j;
long long v[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout.precision(10);
  long long n, first;
  cin >> n >> first;
  long long s = 0;
  for (i = 1; i <= n; i++) cin >> v[i], s += v[i];
  long long minv = 1e15;
  map<long long, long long> cnt;
  for (i = 1; i <= n; i++) {
    cnt[s - v[i]]++;
    minv = min(minv, s - v[i]);
  }
  while (cnt[minv] % first == 0) {
    cnt[minv + 1] += cnt[minv] / first;
    minv++;
  }
  cout << binpow(first, min(minv, s));
  return 0;
}
