CodeForces

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).

Examples

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.

题意是给你一个N位数，和一个素数X，求按公式相加后的最大公约数

引用以下这个博客里的话点击打开链接

分母的最终形式为x^s,s=a1+...+an;

每个分子为x^(s-ai);

如果出现相同分子，则cnt记录一下；

如果出现不同分子，但cnt可以整除x，则前面的幂数加一，i--，将变化后的值与后者相比较

最后用快速幂求分子是多少

#include <iostream>#include <iomanip>#include<stdio.h>#include<string.h>#include<stack>#include<stdlib.h>#include<queue>#include<map>#include<math.h>#include<algorithm>#include<vector>#define inf 1000000007#define LL long long int#define mem(a,b) memset(a,b,sizeof(a))using namespace std; LL a[100005];LL solve(LL x,LL m){    //快速幂    LL ans=1;    x=x%inf;    while(m)    {    if(m&1)     ans=(ans*x)%inf;     x=(x*x)%inf;     m/=2;    }    return ans%inf;}int main(){    LL n,x;    LL s=0;    scanf("%lld%lld",&n,&x);    for(int i=0;i<n;i++)    {        scanf("%lld",&a[i]);        s+=a[i];    }    for(int i=0;i<n;i++)    {        a[i]=s-a[i];//求出分子的幂数    }    a[n]=-1;    //sort排序幂数从小到大    sort(a,a+n);    LL cnt=1,ans;    for(int i=1;i<=n;i++)    {        if(a[i]!=a[i-1])        {            //如果系数可以整除x，则在前者+1，前者值发生改变，i--再判断一次            if(cnt%x==0)            {                cnt/=x;                a[i-1]+=1;                 i--;            }            else            {                //不相等，且系数不能整除x，即为所求                ans=a[i-1];                break;            }        }        else            //如果相等，记录加一            cnt++;    }    ans=min(ans,s);        printf("%lld\n",solve(x,ans));}