Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.

Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.

Input

The first line contains integer q (1?≤?q?≤?1000), the number of handle change requests.

Next q lines contain the descriptions of the requests, one per line.

Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old

The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.

Output

In the first line output the integer n — the number of users that changed their handles at least once.

In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.

Each user who changes the handle must occur exactly once in this description.

Examples

5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov

3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123


題目意思：
有N個改名的動作，輸出改完名的最終結果。

分析：
利用map將key作為新名字，value作為舊名字，使其一一對應
註意好好體會map的用法
自己也看了一會才看明白
code：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<queue>
#include<set>
#include<map>
#include<string>
using namespace std;
typedef long long LL;
int main()
{
    //利用map將key作為新名字，value作為舊名字，使其一一對應。
    string str1,str2;
    int n;
    cin>>n;
    map<string,string>mm;
    map<string,string>::iterator it;
    while(n--)
    {
        cin>>str1>>str2;
        if(mm.find(str1)==mm.end())//如果舊名字不存在，那麽直接將這一對存儲在map中
        {
            mm[str2]=str1;
        }
        else
        {
            //如果就名字存在，那麽將當前的新名字和舊名字所對應的更舊的名字作為一對存儲在map中
            mm[str2]=mm[str1];
            mm.erase(str1);//通過key刪除
        }
    }
    int cont = mm.size();
    cout << cont << endl;
    for(it=mm.begin(); it!=mm.end(); it++)
    {
        cout << it->second << " " << it->first << endl;

    }
    return 0;
}

CodeForces 501B Misha and Changing Handles（STL map）