利用MAP存数据，防止超时，一开始用了两个for 循环CodeForces - 702B

#include<bits/stdc++.h>

using namespace std;

map <int,int> M;

long long ans;

int main()
{
    int n,a;
    cin>>n;
    while(n--)
    {
        cin>>a;
        for(int i=0;i<=31;i++)
            ans+=M[(1LL<<i)-a];
        M[a]++;
    }
    cout<<ans;
}

 CodeForces - 651C 

 

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 .

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

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.

这题用map存，pair为关键字，建立一个m,存(x,y)相等的数目，然后另外两个map1,map2,分别存x相等和y相等的数目，然后map1+map2-m就是答案

#include<iostream>
#include<map>
using namespace std;
typedef long long ll;
map<ll ,ll>map1;
map<ll,ll >map2;
map<pair<long long,long long >,long long >m;
int main()
{
    ll n,x,y;
    ll sum,ans;
    ll i;
    cin>>n;
    map1.clear();
    map2.clear();
    m.clear();
    sum=0,ans=0;
    for(i=0;i<n;i++){
        cin>>x>>y;
        sum+=m[make_pair(x,y)];
        m[make_pair(x,y)]++;
        if(map1.find(x)==map1.end()){
            map1[x]=1;
        }
        else{
            ans+=map1[x];
            map1[x]++;
        }
        if(map2.find(y)==map2.end()){
            map2[y]=1;
        }
        else{
            ans+=map2[y];
            map2[y]++;
        }
    }
    cout<<ans-sum<<endl;
    return 0;
}