codeforces 650 C. Watchmen(数学公式)

C. Watchmen

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi).

They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen i and j to be |xi - xj| + |yi - yj|. Daniel, as an ordinary person, calculates the distance using the formula .

The success of the operation relies on the number of pairs (i, j) (1 ≤ i < j ≤ n), such that the distance between watchman i and watchmen j calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.

Input

The first line of the input contains the single integer n (1 ≤ n ≤ 200 000) — the number of watchmen.

Each of the following n lines contains two integers xi and yi (|xi|, |yi| ≤ 109).

Some positions may coincide.

Output

Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.

Examples

input

3
1 1
7 5
1 5

output

2

input

6
0 0
0 1
0 2
-1 1
0 1
1 1

output

11

Note

In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and  for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.

思路：(|i.x-j.x|+|i.y-j.y|)^2=(i.x-j.x)^2+(i.y-j.y)^2+2*|i.x-j.x|*|i.y-j.y|=(i.x-j.x)^2+(i.y-j.y)^2;

　　　　所以满足条件i.x==j.x或i.y==j.y至少一个;两次分别对x和y排序找到相同的有多少,最后再减去x和y同时相同的就是结果了,因为当时没用long long 导致最后测数据的时候wa了，大哭一场啊！！！

AC代码:

#include <bits/stdc++.h>
using namespace std;
const int N=2e5+6;
struct node
{
    int x,y;
};
node a[N];
int cmp1(node u,node v)
{
    return u.x<v.x;
}
int cmp2(node u,node v)
{
    if(u.y==v.y)return u.x<v.x;
    return u.y<v.y;
}
/*
int cmp3(node u,node v)
{
    if(u.x==v.x)
    {
        return u.y<v.y;
    }
    return u.x<v.x;
}*/
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&a[i].x,&a[i].y);
    }
    sort(a,a+n,cmp1);
    long long num=1,ans=0;
    for(int i=1;i<n;i++)
    {
        if(a[i].x==a[i-1].x)
        {
            num++;
        }
        else
        {
            long long s=(num-1)*num/2;
            ans+=s;
            num=1;
        }
    }
    long long s=(num-1)*num/2;
    ans+=s;
    sort(a,a+n,cmp2);
    num=1;
    for(int i=1;i<n;i++)
    {
        if(a[i].y==a[i-1].y)
        {
            num++;
        }
        else
        {
            long long s=(num-1)*num/2;
            ans+=s;
            num=1;
        }
    }
    s=num*(num-1)/2;
    ans+=s;
    num=1;
    for(int i=1;i<n;i++)
    {
        if(a[i].x==a[i-1].x&&a[i].y==a[i-1].y)
        {
            num++;
        }
        else
        {
            long long s=num*(num-1)/2;
            ans-=s;
            num=1;
        }
    }
            s=num*(num-1)/2;
           ans-=s;

    cout<<ans<<"\n";

    return 0;
}