#F - F

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 (ibi+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⋅1052≤n≤2⋅105) — the number of topics.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤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,…,bnb1,b2,…,bn (1≤bi≤1091≤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
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
这一题还是比较简单的，大概题意就是第一行给你一个数字，代表接下来两行有多少个数字，然后让你寻找下面两行里a[i]+a[j]>b[i]+b[j]的值有多少个。首先我们可以发现，他所求的就是a[i]-b[i]+(a[j]-b[j])>0,所以我们可以把两行里每一项对应的差值算出来，算出差之后，我们就可以将他们排序，我们每次都让最前面的差值与最后面的差值相加，如果>0，那么中间的项都符合要求，然后让最后的项变成-1项，若是不可行，那就让最前面的项+1，这样就可以算出来了。
下面是代码部分：

#include
#include
using namespace std;
int a[210000],b[210000],c[210000];
int main()
{
	int k;
	scanf("%d",&k);
	int i,j,l;
	for(i=0;i<k;i++)
	{
		scanf("%d",&a[i]);
	}
	for(i=0;i<k;i++)
	{
		scanf("%d",&b[i]);
	}
	for(i=0;i<k;i++)
	{
		c[i]=a[i]-b[i];
	}
	sort(c,c+k);
	int n=k-1;
	long long m=0;
	for(i=0;i<n;)
	{
		if(c[i]+c[n]>0)
		{
			m=m+(n-i);
			n=n-1;
		}
		else
		i=i+1;
	}
	printf("%lld",m);
	return 0;
}