codeforce 748B

/*通过map将其对应的字母对应起来，注意：1.该字母是否已经对应过（此题的关键，博主用了3个map，第二个map统计是否已经对应过） 2.相同字母的处理方式（我选择一起处理因为相同字母为不同的一种特例）*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;
int main()
{
string s1, s2;
while(cin >> s1 >> s2){
map <char, char>mp;
map <char, int>mp1;
map <char, int>mp2;
int flag = 0;
int cnt = 0;
for(int i = 0; i < s1.length(); i++){
mp1[s1[i]]++;
mp1[s2[i]]++;
if(s1[i] == s2[i])mp1[s1[i]]--;
if(mp1[s1[i]] != mp1[s2[i]]){
flag = 1;
break;
}
else{
if(mp1[s1[i]] != 1){	//主要是这里如果第二次匹配需要在这里判断然后在对应，不能先对应后判断，那样永远找不到
if(mp[s1[i]] != s2[i] || mp[s2[i]] != s1[i]){
//  cout << s1[i] << " " << s2[i] << endl;
flag = 1;
break;
}
}
mp[s1[i]] = s2[i];
mp[s2[i]] = s1[i];

if(s1[i] != s2[i] && mp1[s1[i]] == 1){
cnt++;

}
}
}
if(flag){
cout << "-1" << endl;
continue;
}
cout << cnt << endl;
for(int i = 0; i < s1.length(); i++){
mp2[s1[i]]++;
mp2[s2[i]]++;
if(s1[i] != s2[i] && mp2[s1[i]] == 1 && mp2[s2[i]] == 1){
cout << s1[i] << " " << s2[i] << endl;

}
}
}
return 0;
}