Watchmen CodeForces - 650A

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原文链接：http://www.cnblogs.com/smallhester/p/10327474.html

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#数据结构与算法

本文解析了CodeForces-650A题目“Watchmen”，介绍了如何计算平面上点对间距离，使用Manhattan距离与Euclidean距离相等的条件，通过高效的数据结构map来统计相同距离的点对数量。

Watchmen CodeForces - 650A

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 (x_i, y_i).

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 |x_i - x_j| + |y_i - y_j|. Daniel, as an ordinary person, calculates the distance using the formula $\text{[math]}$ .

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 x_i and y_i (|x_i|, |y_i| ≤ 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

Output

Input

Output

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 $\text{[math]}$ for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.

题意：给出n个点的坐标(xi,yi)；问有多少对点|xi-xj|+|yi-yj| == sqrt( (xi-xj)^2 + (yi-yj)^2 )。注意：题中有些点的和重合的。

题解：map存存状态，加加减减就好了

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<map>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn =  1e5+5;
const int mod = 1e9+7;

map<ll,ll>mpx;
map<ll,ll>mpy;
map<pair<ll,ll>,ll>mp;
int main()
{
    mpx.clear();
    mpy.clear();
    mp.clear();
    int n;
    ll a,b,ans=0,num=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%lld %lld",&a,&b);
        num+=mp[make_pair(a,b)];
        mp[make_pair(a,b)]++;
        ans+=mpx[a];
        mpx[a]++;
        ans+=mpy[b];
        mpy[b]++;
        //printf("%lld %lld\n",ans,num);
    }
    printf("%lld\n",ans-num);
    return 0;
}