Prime Number

 

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  on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a fraction: , 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 . Thus, the answer to the problem is 8.

In the second sample, . 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 . Thus, the answer to the problem is 1.

 

#include
#include
using namespace std;
 
#define min(a,b) (a>=1;
    }
    return ans;
}
 
int main()
{
    LL x,n,i,sum=0;
    scanf("%lld%lld",&n,&x);
    for(i=0;i

因为x是素数，所以要想使以x为底数的幂指数提高，必须要有x的整数倍数因子。

for循环主要就是为了完成（x^a0+x^a1+……+x^an-1）/(x^sum)    (解释  其中a0