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:
| 題意：給一個字符串和m次操作，每次操作交換ab字符，輸出最後的字符串 |
思路（哈希）：
交換ab字符，可以看成在字典序中的順序改變。如交換字符a（97）和b（98）之後，ASCII 97的位置就變成了b，98的位置就變成了a。這個時候如果說再交換a和c，就找到a的位置（98），c的位置（99），交換，得到a（99）和c（98）。那麼，如果有個字符串abc，在經過上述交換之後，a這個位置本來是97，現在97這個位置變成了b，同理b變成了c，c變成了b，輸出bca 。
AC代碼：
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 3e4+2;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
int Map[500]; // Map記錄字典序對應的字符
int to[500]; //to記錄當前字符對應在Map裏的下標（位置）
int main()
{
ll n; cin>>n; ll m; cin>>m;
string s;
cin>>s;
rep(i,'a','z') Map[i] = i, to[i] = i;
rep(i,1,m)
{
string s1, s2; cin>>s1>>s2;
int pos1 = to[s1[0]], pos2 = to[s2[0]];
Map[pos2] = s1[0];
Map[pos1] = s2[0];
to[s2[0]] = pos1 , to[s1[0]] = pos2;
}
for(int i=0;i<s.size();i++)
s[i] = Map[s[i]];
cout<<s<<'\n';
return 0;
}