Codeforces 591 B：Rebranding

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 lettersxi and yi.

Output

Print the new name of the corporation.

Sample test(s)

input
6 1
police
p m

output
molice

input
11 6
abacabadaba
a b
b c
a d
e g
f a
b b

output
cdcbcdcfcdc

Note

In the second sample the name of the corporation consecutively changes as follows:

$\text{[math]}$

一个公司有一个自己的起始商标，然后它请来了m位设计师，每位设计师都会把之前的商标中，字母x换成字母y，把字符y换成字母x。问经过了m位设计师的折腾之后，最终的商标是什么。

这题一开始想麻烦了。

一共就26个字符，一开始每一个字母都只对应自己，然后每一位设计师来了就change一下，循环每一个字母，看哪一个是设计师要change的那一个，change掉，注意别弄重复了。

代码：

#include <iostream>  
#include <algorithm>  
#include <cmath>  
#include <vector>  
#include <string>  
#include <cstring>  
#pragma warning(disable:4996)  
using namespace std;

int len, m;
char ha[200];
int flag[200];
char st[200005];

void init()
{
    int i;
    for (i = 0; i < 30; i++)
    {
        ha[i] = i;
    }
}

int main()
{
    //freopen("i.txt", "r", stdin);
    //freopen("o.txt", "w", stdout);

    init();
    int i, j;
    char s[10], e[10];
    scanf("%d%d", &len, &m);
    scanf("%s", st);

    for (j = 0; j < m; j++)
    {
        scanf("%s %s", &s, &e);

        memset(flag, 0, sizeof(flag));
        for (i = 0; i < 28; i++)
        {
            if (ha[i] == s[0] - 'a'&&flag[i] != 2)
            {
                ha[i] = e[0] - 'a';
                flag[i] = 1;
            }
            if (ha[i] == e[0] - 'a'&&flag[i] != 1)
            {
                ha[i] = s[0] - 'a';
                flag[i] = 2;
            }
        }
    }
    for (i = 0; i < len; i++)
    {
        printf("%c", ha[st[i] - 'a'] + 'a');
    }
    printf("
");
    //system("pause");
    return 0;
}