CF/359/C. Prime Number

C. Prime Number

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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).

Sample test(s)

input

2 22 2

output

8

input

3 31 2 3

output

27

input

2 229 29

output

73741817

input

4 50 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<stdio.h>#include<string.h>#include<map>#include<iostream>#include<algorithm>using namespace std;const int MOD=1000000007;int a[100010];map<long long ,int>m;long long quick_mul(long long num,long long t){    long long ans=1;    long long pt=num%MOD;    while(t)    {        if(t&1) ans=ans*pt%MOD;        pt=pt*pt%MOD;        t>>=1;    }    return ans;}int main(){  //  freopen("in.txt","r",stdin);    int n,k;    while(~scanf("%d%d",&n,&k))    {        m.clear();        long long sum=0;        for(int i=0;i<n;i++)         {             cin>>a[i];             sum+=a[i];         }         for(int i=0;i<n;i++)         {             long long x=sum-a[i];             m[x]++;             while(m[x]==k){   //k个 k^x 相加= 一个k^(x+1);                 m.erase(x);                 m[++x]++;               }         }         long long ans=min(sum,m.begin()->first); //要注意分子可能比分母大的情况！ 不然就一直wa 8         ans=quick_mul(k,ans);         cout<<ans<<endl;    }    return 0;}