Codeforces Round #464 (Div. 2) D. Love Rescue

D. Love Rescue

time limit per test2 seconds
memory limit per test256 megabytes

Problem Description

Valya and Tolya are an ideal pair, but they quarrel sometimes. Recently, Valya took offense at her boyfriend because he came to her in t-shirt with lettering that differs from lettering on her pullover. Now she doesn’t want to see him and Tolya is seating at his room and crying at her photos all day long.

This story could be very sad but fairy godmother (Tolya’s grandmother) decided to help them and restore their relationship. She secretly took Tolya’s t-shirt and Valya’s pullover and wants to make the letterings on them same. In order to do this, for one unit of mana she can buy a spell that can change some letters on the clothes. Your task is calculate the minimum amount of mana that Tolya’s grandmother should spend to rescue love of Tolya and Valya.

More formally, letterings on Tolya’s t-shirt and Valya’s pullover are two strings with same length n consisting only of lowercase English letters. Using one unit of mana, grandmother can buy a spell of form (c1, c2) (where c1 and c2 are some lowercase English letters), which can arbitrary number of times transform a single letter c1 to c2 and vise-versa on both Tolya’s t-shirt and Valya’s pullover. You should find the minimum amount of mana that grandmother should spend to buy a set of spells that can make the letterings equal. In addition you should output the required set of spells.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the length of the letterings.

The second line contains a string with length n, consisting of lowercase English letters — the lettering on Valya’s pullover.

The third line contains the lettering on Tolya’s t-shirt in the same format.

Output

In the first line output a single integer — the minimum amount of mana t required for rescuing love of Valya and Tolya.

In the next t lines output pairs of space-separated lowercase English letters — spells that Tolya’s grandmother should buy. Spells and letters in spells can be printed in any order.

If there are many optimal answers, output any.

Examples

input
3
abb
dad
output
2
a d
b a
input
8
drpepper
cocacola
output
7
l e
e d
d c
c p
p o
o r
r a

Note

In first example it’s enough to buy two spells: (‘a’,’d’) and (‘b’,’a’). Then first letters will coincide when we will replace letter ‘a’ with ‘d’. Second letters will coincide when we will replace ‘b’ with ‘a’. Third letters will coincide when we will at first replace ‘b’ with ‘a’ and then ‘a’ with ‘d’.

---

解题心得：

1. 题意就是通过将一种字母替换成另一种字母，让给你的A，B两个字符串变得相同，要求改变次数最小。
2. 刚开始瞎搞了半天，然后冷静下来想了一下其实就是一个并查集，只有需要改变的才改变，这样得出来的结果一定是最小的。

---

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+100;

int father[220],n;
char s[2][maxn];

int find(int x){
    if(father[x] == x)
        return x;
    return father[x] = find(father[x]);
}

void merge(int a,int b){
    int fa = find(a);
    int fb = find(b);
    father[fa] = fb;
}

int main(){
    scanf("%d",&n);
    scanf("%s%s",s[0],s[1]);

    for(int i=0;i<210;i++)
        father[i] = i;

    int ans = 0;
    for(int i=0;i<n;i++){
        if(find(s[0][i]) != find(s[1][i])){
            ans++;
            merge(s[0][i],s[1][i]);
        }
    }

    printf("%d
",ans);
    for(int i=0;i<210;i++){
        if(father[i] != i)
            printf("%c %c
",i,father[i]);
    }
    return 0;
}