Codeforces Round #327 (Div. 2) B. Rebranding

B. Rebranding

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The name of one small but proud corporation consists of n lowercase English letters. The Corporation has decided to try rebranding — an active marketing strategy, that includes a set of measures to change either the brand (both for the company and the goods it produces) or its components: the name, the logo, the slogan. They decided to start with the name.

For this purpose the corporation has consecutively hired m designers. Once a company hires the i-th designer, he immediately contributes to the creation of a new corporation name as follows: he takes the newest version of the name and replaces all the letters xiby yi, and all the letters yi by xi. This results in the new version. It is possible that some of these letters do no occur in the string. It may also happen that xi coincides with yi. The version of the name received after the work of the last designer becomes the new name of the corporation.

Manager Arkady has recently got a job in this company, but is already soaked in the spirit of teamwork and is very worried about the success of the rebranding. Naturally, he can't wait to find out what is the new name the Corporation will receive.

Satisfy Arkady's curiosity and tell him the final version of the name.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) — the length of the initial name and the number of designers hired, respectively.

The second line consists of n lowercase English letters and represents the original name of the corporation.

Next m lines contain the descriptions of the designers' actions: the i-th of them contains two space-separated lowercase English letters xiand yi.

Output

Print the new name of the corporation.

Sample test(s)

input

6 1policep m

output

molice

input

11 6abacabadabaa bb ca de gf ab b

output

cdcbcdcfcdc

这是一道很水的题目，但是当时就硬生生地超时了，为什么呢，原因就在于输出，如果最后的输出是用printf将数组中的元素一个一个输出的话，就肯定会超时，但是用了printf("%s",ans); 就不会出现超时的情况。后来我又发现了用getchar和putchar输出是最快的，不管中间搞得多复杂，用getchar和putchar输出是肯定没有问题的。

#include <iostream>using namespace std;char t[200000+10];char ans[125];int main(){int i,j,k,m,n;scanf("%d %d",&m,&n);scanf("%s",t);char a, b;for(i=0;i<125;i++)ans[i]=i;for(k=0;k<n;k++){scanf(" %c %c",&a,&b);for(i='a';i<='z';i++){if(ans[i]== a ){ans[i] = b;}else if(ans[i] ==b ){ans[i] = a;}}}for(j=0;j<m;j++)        t[j]=ans[t[j]];printf("%s",t);return 0;}