Prime Number CodeForces - 359C (属于是数论)

Simon has a prime number x and an array of non-negative integers a₁, a₂, ..., a_n.

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 x^{a₁ + a₂ + ... + a_n}.
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, ..., a_n (0 ≤ a₁ ≤ a₂ ≤ ... ≤ a_n ≤ 10⁹).

Output

Print a single number — the answer to the problem modulo 1000000007 (10⁹ + 7).

Example

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,
as351 = 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<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

const long long INF=1e9+7;
const long long maxn=101000;

long long n,x;
long long a[maxn];

long long quick_mod(long long a,long long b)
{
    long long ans=1;
    a=a%INF;
    while(b)
    {
        if(b&1)
            ans=ans*a%INF;
        a=a*a%INF;
        b>>=1;
    }
    return ans;
}

int main()
{
    while(~scanf("%d %d",&n,&x))
    {
        long long sum1=0;
        for(long long i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
            sum1+=a[i];
        }
        for(long long i=0; i<n; i++)
            a[i]=sum1-a[i];
        sort(a,a+n);
        long long ans,j=1,cot=1,t;
        for(j=1; j<=n; j++)
        {
            if(a[j]!=a[j-1])
            {
                if(cot%x)
                {
                    ans=a[j-1];
                    break;
                }
                else
                {
                    cot/=x;
                    a[j-1]+=1;
                    j--;
                }
            }
            else cot++;
        }
        printf("%d\n",quick_mod(x,min(ans,sum1)));
    }
    return 0;
}

#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

const long long INF=1e9+7;
const long long maxn=101000;

long long n,x;
long long a[maxn];

long long gcd(long long a,long long b)
{
    long long ans=1;
    ans=ans%INF;
    while(b)
    {
        if(b&1)
            ans=ans*a%INF;
        a=a*a%INF;
        b>>=1;
    }
    return ans;
}

int main()
{
    while(~scanf("%I64d %I64d",&n,&x))
    {
        long long sum1=0;
        for(long long i=0; i<n; i++)
        {
            scanf("%I64d",&a[i]);
            sum1+=a[i];
        }
        for(long long i=0; i<n; i++)
            a[i]=sum1-a[i];
        sort(a,a+n);
        long long ans,j=1,cot=1,t;
        while(j<=n)
        {
            if(a[j]!=a[j-1])
            {
                if(cot%x)
                {
                    ans=a[j-1];
                    break;
                }
                long long f=a[j-1]+1;
                t=cot/x;
                for(long long k=j-1,s=t; s>0; s--,k--)
                    a[k]=f;
                j-=t;
                j++;
                cot=1;
            }
            else cot++,j++;
        }
        printf("%I64d\n",gcd(x,min(ans,sum1)));
    }
    return 0;
}