A

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 x_i by y_i, and all the letters y_i by x_i. 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 x_i coincides with y_i. 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 x_i and y_i.

Output

Print the new name of the corporation.

Example

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:

题目意思:给一串字符串,再给出一系列指令 如 a b ;就是把现在字符串中的a->b,b->a;

解题思路:可以读取一次指令转换一次,不过这样一定会超时,所以先读取完指令后,可以得到a最后会转换成什么;

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;

typedef long long ll;
const double T = 3.141592653589793;
const int MAX = 2000000;
char a[MAX];

int main()
{
    int N,n;
    cin>>N>>n;
    int visit[30] = {0};
    for(int i = 1;i<=N;i++)
       {
           cin>>a[i];
           int temp = a[i] -'a'+1;
           if(visit[temp]==0)
                visit[temp] = temp;
       }


    while(n--)
    {
        char t1,t2;
        cin>>t1>>t2;
        int t1x = t1 - 'a'+1;
        int t2x = t2 -'a'+1;
        for(int i = 1;i<=26;i++)
        {
            if(visit[i]==t1x)
                visit[i] = t2x;
            else if(visit[i]==t2x)
                visit[i] = t1x;
        }
    }

    for(int i = 1;i <=N;i++)
    {
        int temp = a[i] -'a'+1;
        char T = visit[temp] - 1+'a';
        cout<<T;
    }
    cout<<endl;


    return 0;
}