Little A gets to know a new friend, Little B, recently. One day, they realize that they are family 500 years ago. Now, Little A wants to know whether Little B is his elder, younger or brother.

There are multiple test cases.

For each test case, the first line has a single integer, n (n<=1000). The next n lines have two integers a and b (1<=a,b<=2000) each, indicating b is the father of a. One person has exactly one father, of course. Little A is numbered 1 and Little B is numbered 2.

Proceed to the end of file.

For each test case, if Little B is Little A’s younger, print “You are my younger”. Otherwise, if Little B is Little A’s elder, print “You are my elder”. Otherwise, print “You are my brother”. The output for each test case occupied exactly one line.

#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

int pre[];

void Solve(){

    int cnta = , cntb = , a = , b = ;

    while(pre[a]){

        cnta += ;

        a = pre[a];

    }

    while(pre[b]){

        cntb += ;

        b = pre[b];

    }

    if(cnta == cntb) printf("You are my brother\n");

    else printf("%s\n", cnta < cntb?"You are my younger":"You are my elder");

}

int main(){

    int n;

    while(scanf("%d", &n) == ){

        int a, b;

        memset(pre, , sizeof(pre));

        for(int i = ; i < n; i++){

            scanf("%d %d", &a, &b);

            pre[a] = b;

        }

        Solve();

    }

}