Prime Number CodeForces

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 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, 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;}