codeforces 1324 D. Pair of Topics

The next lecture in a high school requires two topics to be discussed. The ii-th topic is interesting by aiai units for the teacher and by bibi units for the students.

The pair of topics ii and jj (i<ji<j) is called good if ai+aj>bi+bjai+aj>bi+bj (i.e. it is more interesting for the teacher).

Your task is to find the number of good pairs of topics.

Input

The first line of the input contains one integer nn (2≤n≤2⋅105) — the number of topics.

The second line of the input contains nn integers a1,a2,…,an (1≤ai≤109), where aiai is the interestingness of the ii-th topic for the teacher.

The third line of the input contains nn integers b1,b2,…,bn (1≤bi≤109), where bibi is the interestingness of the ii-th topic for the students.

Output

Print one integer — the number of good pairs of topic.

Examples

input

Copy

5
4 8 2 6 2
4 5 4 1 3

output

Copy

7

input

Copy

4
1 3 2 4
1 3 2 4

output

Copy

0

解题思路：

  由于 n 的大小限制，两个for循环必然超时。

 调整不等式  为  (Ai - Bi) + (Aj - Bj)  > 0 ，那么此时就可将两个数组转换为一个数组

然后从小到大排序，利用二分，对于第i项，通过二分在[i+1,n]找到最小的j，满足该不等式，使用upper_bound函数即可

#include<bits/stdc++.h>

using namespace std;

const int N = 1e6+10;

int a[N],b[N],c[N],d[N];

int main()

{

    std::ios::sync_with_stdio(false);

    int n;

    cin>>n;

    for(int i = 1; i <= n; i++) {

        cin>>a[i];

    }

    for(int i = 1; i <= n; i++) {

        cin>>b[i];

        c[i] = a[i]-b[i];

    }

    sort(c+1,c+n+1);

    long long ans=0;

    for(int i=1;i<=n;i++){

        int pos=upper_bound(c+i+1,c+n+1,-c[i])-c-1;

        if(pos>=n+1)

            continue;

        ans += (n-pos);

    }

    cout<<ans<<endl;

    return 0;

}