count given N points how many squares parallel to XY axis ca

/* count given N points how many squares parallel to XY axis can b formed...Here if one x point has more than sqrt(N) points the Y axis of them will have less than sqrt(N) points...codeforces 425D */

vi X[MAX],Y[MAX];ll ans;

set<int>XX[MAX],YY[MAX];

void dfs(int s,int flag){
    int len;
    if(flag){
        len = X[s].size();
        for(int i = 0;i<len;i++){
            for(int j = i+1;j<len;j++){
                int x = X[s][i];
                int xx = X[s][j];
                int d = abs(x - xx);

                if(s - d>=0&&s - d<MAX&&XX[s - d].find(x)!=XX[s-d].end()&&XX[s-d].find(xx)!=XX[s-d].end())ans++;
                if(s + d>=0&&s + d<MAX&&XX[s + d].find(x)!=XX[s+d].end()&&XX[s+d].find(xx)!=XX[s+d].end())ans++;
            }
        }
    }
    else {
        len = Y[s].size();
         for(int i = 0;i<len;i++){
            for(int j = i+1;j<len;j++){
                int x = Y[s][i];
                int xx = Y[s][j];
                int d = abs(x - xx);

                if(s - d>=0&&s - d<MAX&&YY[s - d].find(x)!=YY[s-d].end()&&YY[s-d].find(xx)!=YY[s-d].end())ans++;
                if(s + d>=0&&s + d<MAX&&YY[s + d].find(x)!=YY[s+d].end()&&YY[s+d].find(xx)!=YY[s+d].end())ans++;
            }
        }
    }
}


int main()
{
    booster()
    ///read("input.txt");

    int n;
    cin>>n;
    for(int i = 1;i<=n;i++){
        int x,y;
        cin>>x>>y;
        XX[x].insert(y);
        X[x].pb(y);
    }

    int sz = sqrt(n);

    for(int i = 0;i<MAX;i++){
        if(X[i].size()>sz){
            for(auto it : X[i]){
                YY[it].insert(i);
                Y[it].pb(i);
            }
        }
        else {
            dfs(i,1);
            XX[i].clear();
            X[i].clear();
        }
    }
    for(int i = 0;i<MAX;i++){
        dfs(i,0);
        YY[i].clear();
        Y[i].clear();
    }

    cout<<ans;

    return 0;
}