CodeForces 651C Watchmen map

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

 题目大意：

给你很多组a,b  然后通过两种计算方法判断答案是否相同

1.  |xi-xj|+|yi-yj|

2. sqrt((xi-xj)*(xi-xj) + (yi-yj)*(yi-yj))

 

题解：发现只有当xi = xj 或 yi = yj时答案相同

但同时要排除(xi,yi)=(xj,yj)的情况

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
map<int, int> mapx, mapy;
map<pair<int, int>, int> mapp;
int main()
{
    int n, x, y;
    scanf("%d", &n);
    mapx.clear();
    mapy.clear();
    mapp.clear();
    for (int k = 1; k <= n; k++)
    {
        scanf("%d%d", &x, &y);
        mapx[x]++;
        mapy[y]++;
        mapp[pair<int, int>(x, y)]++;
    }
    ll ans = 0;
    map<int, int>::iterator i;
    int tem;
    for (i = mapx.begin(); i != mapx.end(); i++)
    {
        tem = i->second;
        ans += (ll)tem * (tem - 1) / 2;
    }
    for (i = mapy.begin(); i != mapy.end(); i++)
    {
        tem = i->second;
        ans += (ll)tem * (tem - 1) / 2;
    }
    for (map<pair<int, int>, int>::iterator j = mapp.begin(); j != mapp.end(); j++)
    {
        tem = j->second;
        ans -= (ll)tem * (tem - 1) / 2;
    }
    printf("%I64d\n", ans);
    return 0;
}