Problem:
Score : 200 points

Problem Statement

Takahashi is practicing shiritori alone again today.

Shiritori is a game as follows:

In the first turn, a player announces any one word.

In the subsequent turns, a player announces a word that satisfies the following conditions:

That word is not announced before.

The first character of that word is the same as the last character of the last word announced.

In this game, he is practicing to announce as many words as possible in ten seconds.

You are given the number of words Takahashi announced, N, and the i-th word he announced, W_i, for each i. Determine if the rules of shiritori was observed, that is, every word announced by him satisfied the conditions.

Constraints

N is an integer satisfying 2 \leq N \leq 100.

W_i is a string of length between 1 and 10 (inclusive) consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
W_1
W_2
:
W_N

Output

If every word announced by Takahashi satisfied the conditions, print Yes; otherwise, print No.

Sample Input 1

4
hoge
english
hoge
enigma

Sample Output 1

No

As hoge is announced multiple times, the rules of shiritori was not observed.

Sample Input 2

9
basic
c
cpp
php
python
nadesico
ocaml
lua
assembly

Sample Output 2

Yes

Sample Input 3

8
a
aa
aaa
aaaa
aaaaa
aaaaaa
aaa
aaaaaaa

Sample Output 3

No

Sample Input 4

3
abc
arc
agc

Sample Output 4

No

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(){
    int N;
    int a=0;
    char W[128][16];
    scanf("%d",&N);
    for(int i = 0; i<N; i++){
        scanf("%s",W[i]);
    }
    for(int i = 0; i<N; i++){
        int x=0;
        for(int j = 0; j<N; j++){
            int y=0;
            for(int n = 0; n<16; n++){
                if (W[i][n]==W[j][n]){y++;}
            }
            if(y==16){++x;}
        }
        if(x==1){}else{a=1;}
    }
    
    
    int x=0;
    for(int i = 0; i<N; i++){
        int k=0;
        for(int j = 0; j<16; j++){
            k=j;
            if(W[i][j]==0){break;}
        }
        if(W[i][k-1]==W[i+1][0]){}else{
            ++x;
        }
    }
    if(x==1){}else{a=1;}
    
    if(a==0){printf("Yes");}
    if(a==1){printf("No");}
    return 0;
}