周赛--Heros and Swords

There are n swords of different weights Wiand n heros ofpower
Pi.

Your task is to find out how many ways the heros can carry the swords so that each hero carries exactly one sword.



Here are some rules:

(1) Every sword is carried by one hero and a hero cannot carry a sword whose weight is larger than his power.
(2) Two ways will be considered different if at least one hero carries a different sword.

Input

The first line of the input gives the number of test cases T(1 ≤ T ≤ 50).

Each case starts with a line containing an integer n (1 ≤ n ≤ 105)
denoting the number of heros and swords.

The next line contains n space separated distinct integers denoting the weight of swords.

The next line contains n space separated distinct integers denoting the power for the heros.

The weights and the powers lie in the range [1, 109].

Output

For each case, output one line containing "Case #x: " followed by the number of ways those heros can carry the swords.

This number can be very big. So print the result modulo 1000 000 007.

Sample Input

3
5
1 2 3 4 5
1 2 3 4 5
2
1 3
2 2
3
2 3 4
6 3 5

Sample Output

Case #1: 1
Case #2: 0
Case #3: 4

#include <iostream>
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <set>
#include <stdio.h>
#include <time.h>
using namespace std;
#define LL long long
#define pi acos(-1.0)
#pragma comment(linker, "/STACK:1024000000")
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
const double eqs=1e-9;
const int MAXN=1e5+10;
int a[MAXN], b[MAXN];
int main()
{
        int t, n, i, j, num=0;
        LL ans;
        scanf("%d",&t);
        while(t--){
                scanf("%d",&n);
                for(i=0;i<n;i++){
                        scanf("%d",&a[i]);
                }
                for(i=0;i<n;i++){
                        scanf("%d",&b[i]);
                }
                sort(a,a+n);
                sort(b,b+n);
                ans=1;
                j=0;
                for(i=0;i<n;i++)
            {
                        while(j<n&&a[j]<=b[i])
                        {
                                j++;
                        }
                        ans=ans*(j-i)%mod;
                }
                printf("Case #%d: %lld\n",++num,ans);
        }
        return 0;
}