The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by ai units for the teacher and by bi units for the students.
The pair of topics i and j (i<j) is called good if ai+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 n (2≤n≤2⋅105) — the number of topics.
The second line of the input contains n integers a1,a2,…,an (1≤ai≤109), where ai is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b1,b2,…,bn (1≤bi≤109), where bi is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0
题目大意：给出两个数组，求ai+aj>bi+bj的种数。
解题思路：ai+aj>bi+bj可以转化为ai-bi+aj-bj>0的种数。我们开三个数组，一个用于存放a，一个用于存放b，最后一个数组c用来存放ai-bi的值。我们给c数组从小到大排个序，每次查找比 -ci 大的值，这样ci+cj一定大于0。
PS：其实不用考虑 i<j 因为 i ！= j 所以随便找出两个数，他们的编号必须是一个大一个小。AC代码：
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int _max=2e5+50;
using LL = long long;
LL a[_max],b[_max],c[_max];
int main()
{
int n;
cin>>n;
for(int i=0;i<n;i++)
cin>>a[i];
for(int i=0;i<n;i++)
{
cin>>b[i];
c[i]=a[i]-b[i];
}
sort(c,c+n);
LL ans=0;
for(int i=0;i<n;i++)
{
int x=upper_bound(c+i+1,c+n,-c[i])-c;
if(x>=n)//如果找不到返回最后一个地址+1
continue;
else
ans+=n-x;//因为一共n个数，所以用n-x就得到数量了
}
cout<<ans<<endl;
//system("pause");
return 0;
}