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;
using cd = complex<double>;
const long long modd1 = 1e9 + 7, modd2 = 998244353, maxn = 3e5 + 5, K = 26,
                inf = 1e9, infll = 1e18;
const double pi = acos(-1);
map<long long, long long> occ;
long long a[maxn];
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % modd1;
    b >>= 1;
    a = (a * a) % modd1;
  }
  return res;
}
void solve() {
  long long n, i, j, k, l, r, m, sum = 0, x, ans;
  cin >> n >> x;
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    sum += a[i];
  }
  for (i = 1; i <= n; ++i) {
    a[i] = sum - a[i];
    occ[a[i]]++;
  }
  for (auto v : occ) {
    long long ap = v.second;
    if (ap % x != 0) {
      ans = binpow(x, min(sum, v.first));
      cout << ans;
      break;
    }
    long long cnt = 0;
    while (ap % x == 0) {
      cnt++;
      ap /= x;
    }
    occ[v.first] = 0;
    occ[v.first + cnt] += ap;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}
