CodeForces 501B Misha and Changing Handles（STL map）

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 and new are distinct. The lengths of the strings do not exceed 20.

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

Input

5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov

Output

3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123

题目意思：
有N个改名的动作，输出改完名的最终结果。

分析：
利用map将key作为新名字，value作为旧名字，使其一一对应
注意好好体会map的用法
自己也看了一会才看明白
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将key作为新名字，value作为旧名字，使其一一对应。
    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())//如果旧名字不存在，那么直接将这一对存储在map中
        {
            mm[str2]=str1;
        }
        else
        {
            //如果就名字存在，那么将当前的新名字和旧名字所对应的更旧的名字作为一对存储在map中
            mm[str2]=mm[str1];
            mm.erase(str1);//通过key删除
        }
    }
    int cont = mm.size();
    cout << cont << endl;
    for(it=mm.begin(); it!=mm.end(); it++)
    {
        cout << it->second << " " << it->first << endl;

    }
    return ;
}