Codeforces Round #345 (Div. 2) C. Watchmen __ map , sorting and combinatorics

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.

Source

Codeforces Round #345 (Div. 2) C. Watchmen

My Solution

build a 

struct point{
    int x, y;
} pts[maxn];

to store the data, and sort with .x first and to calculate the same x that is on continuouslyand use C(2, t) = t*(t-1)/2 add to the sumx;

and sumy is the same way to do as the sumx did.

Then I use a map<int,map<int, int> > cnt to count the same coordinates when input, and them use iterator to count the sumsame;

All of sumx, sumy, sumsame, t, num should be long long, because the t,num can be 2*1e5 so the C(2, t) and C(2, num) can be 400 1e8 large.

#include <iostream>
#include <cstdio>
#include <map>
//#include <set>
#include <algorithm>
using namespace std;
const int maxn = 200000 + 8;

struct point{
    int x, y;
} pts[maxn];

bool cmpx(const point &a, const point &b)
{
    return a.x < b.x;
}

bool cmpy(const point &a, const point &b)
{
    return a.y < b.y;
}

map<int,map<int, int> > cnt;

int main()
{
    //coordinate is not the same as the edges
    //and WA28 because t and num should be long long ,too    the val can be 200000  t*(t-1) num*(num-1) canbe 40 000 000 000
    #ifdef LOCAL
    freopen("a.txt", "r", stdin);
    #endif // LOCAL
    int n;
    long long sumx = 0, sumy = 0, sumsame = 0;
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%d%d", &pts[i].x, &pts[i].y);
        cnt[pts[i].x][pts[i].y]++;
    }
    //the same x
    sort(pts, pts+n, cmpx);
    long long t = 1;
    int val = pts[0].x;
    for(int i = 1; i < n; i++){
        if(val == pts[i].x) t++;
        else {val = pts[i].x; sumx += (t*(t-1))/2; t = 1;}
    }
    //the last will be calculated out the for body
    sumx += (t*(t-1))/2;

    //the same y
    sort(pts, pts+n, cmpy);
    t = 1;
    val = pts[0].y;                                        //printf("%d %d\n", pts[0].x, pts[0].y);
    for(int i = 1; i < n; i++){
        //printf("%d %d\n", pts[i].x, pts[i].y);
        if(val == pts[i].y) t++;
        else {val = pts[i].y; sumy += (t*(t-1))/2; t = 1;}
    }
    //the last will be calculated out the for body
    sumy += (t*(t-1))/2;

    //Some positions may coincide.
    long long num;
    for(auto i = cnt.begin(); i != cnt.end(); i++){
        for(auto j = (i->second).begin(); j != i->second.end(); j++){
            num = j->second;
            if(num > 1) sumsame += (num*(num-1))/2;
        }
    }
    //printf("%lld %lld %lld\n",sumx, sumy, sumsame);
    printf("%I64d", sumx+sumy-sumsame);
    return 0;
}

Thank you all!