Prime Number

最新推荐文章于 2021-08-16 00:31:41 发布

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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

Input

3 3
1 2 3

Output

Input

2 2
29 29

Output

73741817

Input

4 5
0 0 0 0

Output

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<algorithm>
using namespace std;
 
#define min(a,b) (a<b?a:b)
#define LL long long int
#define maxn 100000+10
#define g 1000000007
LL a[maxn];
 
LL quick_mod(LL a,LL b)
{
    LL ans=1;
    a=a%g;
    while(b)
    {
        if(b&1)
            ans=ans*a%g;
        a=a*a%g;
        b>>=1;
    }
    return ans;
}
 
int main()
{
    LL x,n,i,sum=0;
    scanf("%lld%lld",&n,&x);
    for(i=0;i<n;i++)
    {
        scanf("%lld",&a[i]);
        sum+=a[i];//分母的次数k
    }
    for(i=0;i<n;i++)
        a[i]=sum-a[i];//分子的次数k
    sort(a,a+n);
    LL ans,cnt=1;//cnt为系数
    a[n]=-1;
    for(i=1;i<=n;i++)
    {
        if(a[i]!=a[i-1])
        {
            if(cnt%x)
            {
                ans=a[i-1];
                break;
            }
            else
            {
                cnt/=x;
                a[i-1]+=1;
                i--;
            }
        }
        else cnt++;
    }
    ans=min(ans,sum);
    printf("%lld\n",quick_mod(x,ans));
    return 0;
}

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<a1<……<an-1) 当分子分母同时除以必有的公因子x^a0后。得到( 1+x^(a1-a0)+x^(a2-a0)+……+x^(an-1-a0) )/( x^(sum-a0) ) 如果此时（指数从小到大看）有指数一样的就合并，若合并的结果是x的整数倍，那么x的指数就增大一，则公因子就增大x倍。巧妙的是，指数相等合并并不是总体的，从小到大一段一段来的。