You are given a permutation a from 0 to n−1 and a permutation b from 0 to m−1.

Define that the domain of function f is the set of integers from 0 to n−1, and the range of it is the set of integers from 0 to m−1.

Please calculate the quantity of different functions f satisfying that f(i)=bf(ai) for each i from 0 to n−1.

Two functions are different if and only if there exists at least one integer from 0 to n−1 mapped into different integers in these two functions.

The answer may be too large, so please output it in modulo 109+7.

 

The input contains multiple test cases.

For each case:

The first line contains two numbers n, m. (1≤n≤100000,1≤m≤100000)

The second line contains n numbers, ranged from 0 to n−1, the i-th number of which represents ai−1.

The third line contains m numbers, ranged from 0 to m−1, the i-th number of which represents bi−1.

It is guaranteed that ∑n≤106, ∑m≤106.

 

For each test case, output "Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

 

 

 

Source

 

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#include <iostream>

#include<cstdio>

#include<cstring>

#include<queue>

#include<bits/stdc++.h>

using namespace std;

const long long mod=1e9+;

int n,m;

int a[],b[],a1[],b1[];

bool vis[];

int acir,bcir;

int main()

{

    int cas=;

    while(~scanf("%d%d",&n,&m))

    {

        for(int i=;i<n;i++) scanf("%d",&a[i]);

        for(int i=;i<m;i++) scanf("%d",&b[i]);

        memset(vis,,sizeof(vis));

        acir=;

        for(int i=;i<n;i++)

        {

            if (vis[i]) continue;

            int k=i;

            acir++; a1[acir]=;

            while(!vis[k])

            {

                vis[k]=;

                a1[acir]++;

                k=a[k];

            }

        }

        bcir=;

        memset(vis,,sizeof(vis));

        for(int i=;i<m;i++)

        {

            if (vis[i]) continue;

            int k=i;

            bcir++; b1[bcir]=;

            while(!vis[k])

            {

                vis[k]=;

                b1[bcir]++;

                k=b[k];

            }

        }

       long long ans=;

       for(int i=;i<=acir;i++)

       {

           long long tmp=;

           for(int j=;j<=bcir;j++)

               if (a1[i]%b1[j]==) tmp+=b1[j];

           ans=ans*tmp%mod;

       }

       printf("Case #%d: %lld\n",++cas,ans%mod);

    }

    return ;

}