Problem: 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 xi by 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 xi and yi.

Output

Print the new name of the corporation.

Examples

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:

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Solution (in JAVA):

Solution:
import java.util.*;
import java.math.*;
import java.io.*;
import java.text.*; 
 public class A{

    static FastReader scan=new FastReader();
   public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
  
static char mat[][]=new char[105][105];

long  get( long x){
    return x*(x-1)/2;
}
static boolean can(int x,int y){
    if(mat[x-1][y]!='#')return false;
    if(mat[x+1][y]!='#')return false;
    if(mat[x][y-1]!='#')return false;
    if(mat[x][y+1]!='#')return false;
    return true;

}
static void make(int x,int y){
    mat[x][y]=mat[x+1][y]=mat[x-1][y]=mat[x][y-1]=mat[x][y+1]='.';

}
public static void main(String[] args)  throws Exception 
{
/*long n=scan.nextLong();
long m=scan.nextLong();
if(m==1)
{
    out.print((n*2)+" "+(n*2));
    out.close();
    return;
}
long min=n/m;
min*=m;
min/=2;
out.print(min+" ");
long max=m-1;
n-=max;
out.println(n*(n-1)/2);*/
//KMP
/*
KMP uses degenerating property(pattern having same sub-
patterns appearing more than once in the pattern)
the idea:whenever we detect a mismatch(after some matches)
,we already know some of the characters in the text of the 
next window.We take advantage of this info to avoid matching 
the chars that we know will anyway match. 

TIME O(N) in the worst case
*/
char change[]=new char[26];
for(int i=0;i<26;i++)
change[i]=(char)(i+'a');
int n=scan.nextInt(),m=scan.nextInt();
StringBuilder s=new StringBuilder(scan.next());
for(int i=0;i<m;i++){
    char c=scan.next().charAt(0),cc=scan.next().charAt(0);
    for(int j=0;j<26;j++)
    {
        if(change[j]==c)
            change[j]=cc;
       else if(change[j]==cc)
            change[j]=c;
    }
}
for(int i=0;i<n;i++)
{
    s.setCharAt(i,change[s.charAt(i)-'a']);
   // out.println(change[s.charAt(i)-'a']);
}
out.println(s);

      
 out.close();








  

}
 public static  long pow(long b, long e) {
            long r = 1;
            while (e > 0) {
                if (e % 2 == 1) r = r * b ;
                b = b * b;
                e >>= 1;
            }
            return r;
        }
private static void sort(long[] arr) {
        List<Long> list = new ArrayList<>();
        for (long object : arr) list.add(object);
        Collections.sort(list);
        for (int i = 0; i < list.size(); ++i) arr[i] = list.get(i);
    }
 public static class FastReader {
        BufferedReader br;
        StringTokenizer root;
        
 
        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        String next() {
            while (root == null || !root.hasMoreTokens()) {
                try {
                    root = new StringTokenizer(br.readLine());
                } catch (Exception addd) {
                    addd.printStackTrace();
                }
            }
            return root.nextToken();
        }
 
        int nextInt() {
            return Integer.parseInt(next());
        }
 
        double nextDouble() {
            return Double.parseDouble(next());
        }
 
        long nextLong() {
            return Long.parseLong(next());
        }
 
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (Exception addd) {
                addd.printStackTrace();
            }
            return str;
        }
         public int[] nextIntArray(int arraySize) {
            int array[] = new int[arraySize];
 
            for (int i = 0; i < arraySize; i++) {
                array[i] = nextInt();
            }
 
            return array;
        }
}
static class pair{
    int i;
    int j;
pair(int i,int j){
    this.i=i;
    this.j=j;
}

}}