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 long long Mod = 1000000007;
long long x, n, sum;
multiset<long long> ms;
multiset<long long>::iterator it_ms;
set<long long> s;
set<long long>::iterator it_s;
map<long long, long long> mp;
inline long long Read() {
  long long num = 0, op = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = num * 10 + ch - '0';
    ch = getchar();
  }
  return num * op;
}
inline long long QuickPower(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % Mod;
    a = a * a % Mod;
    b >>= 1;
  }
  return ans;
}
signed main() {
  n = Read(), x = Read();
  for (long long i = 1; i <= n; ++i) {
    long long v = Read();
    sum += v;
    ms.insert(v);
  }
  for (it_ms = ms.begin(); it_ms != ms.end(); it_ms++) {
    ++mp[sum - *it_ms];
    s.insert(sum - *it_ms);
  }
  for (it_s = s.begin(); it_s != s.end(); it_s++) {
    if (mp[*it_s] >= x) {
      mp[(*it_s) + 1] += mp[*it_s] / x;
      mp[*it_s] %= x;
      ms.insert((*it_s) + 1), s.insert((*it_s) + 1);
    }
  }
  for (it_s = s.begin(); it_s != ms.end(); it_s++) {
    if (mp[*it_s]) {
      printf("%lld\n", QuickPower(x, min(sum, *it_s)));
      return 0;
    }
  }
  return 0;
}
