Misha and Changing Handles

Question

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.

Example

Input

5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov

---

Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123

Code

#include 
#include 
using namespace std;

char a[1005][25],b[1005][25];
int main()
{
    int n,i,j,m=0;
    cin>>n;
    for(i=0;icin>>a[i]>>b[i];
    }
    for(i=0;iif(strlen(a[i])!=0)
        {
            for(j=i+1;jif(strcmp(b[i],a[j])==0)
                {
                    a[j][0]='\0';
                    strcpy(b[i],b[j]);
                }
            }
        }
    }
    for(i=0;iif(strlen(a[i])!=0)
            m++;
    cout<for(i=0;iif(strlen(a[i])!=0)
        {
            cout<' '<return 0;
}

这里我想说明的是”两个字符串之间用空格隔开这一条件“。不同的输入函数对空格的处理是不同的，scanf()和cin遇到空格则当前的字符串输入结束，即空格不放入字符串中，而gets()只有在遇到会车时才输入结束，因此空格会在字符串中，最初使用gets(),则导致写了一个非常复杂的函数，细节往往会影响我们的效率。