A. Bark to Unlock

A. Bark to Unlock

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.

Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark n distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.

Input

The first line contains two lowercase English letters — the password on the phone.

The second line contains single integer n (1 ≤ n ≤ 100) — the number of words Kashtanka knows.

The next n lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.

Output

Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.

You can print each letter in arbitrary case (upper or lower).

Examples

Input

ya
4
ah
oy
to
ha

Output

YES

Input

hp
2
ht
tp

Output

NO

Input

ah
1
ha

Output

YES

Note

In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".

In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.

In the third example the string "hahahaha" contains "ah" as a substring.

这题直接暴力枚举就够了。

#include <bits/stdc++.h>
using namespace std;
char a[3];
string k[105];
int n;
int main(){
  cin>>a;
  cin>>n;
  bool prime=true,flag=true;
  int t=0;
  while(t!=n){
      cin>>k[t];
      t++;
  }
  for(int i=0;i<n;i++){
    if(k[i]==a)
      {
        cout<<"YES"<<endl;
        return 0;
      }
  }
  for(int i=0;i<n;i++){
    if(k[i][1]==a[0]){
      prime=false;
    }
    if(k[i][0]==a[1]){
      flag=false;
    }

  }

  if(!flag&&!prime){
    cout<<"YES"<<endl;
  }else{
    cout<<"NO"<<endl;
  }
  return 0;
}