As a good tuner, DreamGrid can choose an integer $K$ (can be positive, 0, or negative) as his tune and add $K$ to every element in $D$. Can you help him maximize his score by choosing a proper tune?

Input

There are multiple test cases. The first line of the input contains an integer $T$ (about 100), indicating the number of test cases. For each test case:

The first line contains one integer $n$ ($1 \le n \le 10^5$), indicating the length of the sequences $D$ and $S$.

The second line contains $n$ integers $D_1, D_2, \dots, D_n$ ($-10^5 \le D_i \le 10^5$), indicating the song performed by DreamGrid.

The third line contains $n$ integers $S_1, S_2, \dots, S_n$ ($-10^5 \le S_i \le 10^5$), indicating the standard version of the song.

It‘s guaranteed that at most 5 test cases have $n > 100$.

Output

For each test case output one line containing one integer, indicating the maximum possible score.

Sample Input

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

Sample Output

3
1

Hint

For the first sample test case, DreamGrid can choose $K = 1$ and changes $D$ to $\{2,3,4,5\}$.

For the second sample test case, no matter which $K$ DreamGrid chooses, he can only get at most 1 match.

第一个数列各个位上的数字加上一个k可以得到第二个数列

问最多有多少个数可以通过加上数字k等于第二个数列

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <map>

using namespace std ; 

#define maxn 1100000
int num1[maxn] , num2[maxn] , num[maxn] ; 
int t ; 
int n ; 
map<int,int> mp ; 
// 数据太多而且不方便枚举 数字 k 所以使用map
int main(){

    cin >> t ; 

    while(t--){
        cin >> n ; 
        mp.clear() ; 
        for(int i=1 ; i<=n ; i++){
            cin >> num1[i] ; 
        }
        for(int i=1 ; i<=n ; i++){
            cin >> num2[i] ; 
        }

        for(int i=1 ; i<=n ; i++){
            mp[num1[i]-num2[i]] ++ ; 
        }
        int result = 0 ; 
        // 遍历map
        for(map<int,int>::iterator item = mp.begin();item!=mp.end() ;item++){
            if(item->second > result){
                result = item->second ; 
            }
        }
        cout << result << endl ; 
    }
    return 0 ; 
}