Problem Statement

We found a dictionary of the Ancient Civilization Mayo (ACM) during excavation of the ruins. After analysis of the dictionary, we revealed they used a language that had not more than 26 letters. So one of us mapped each letter to a different English alphabet and typed all the words in the dictionary into a computer.

How the words are ordered in the dictionary, especially whether they are ordered lexicographically, is an interesting topic to many people. As a good programmer, you are requested to write a program to judge whether we can consider the words to be sorted in a lexicographical order.

Note: In a lexicographical order, a word always precedes other words it is a prefix of. For example, `ab` precedes `abc`, `abde`, and so on.



Input

The input consists of multiple datasets. Each dataset is formatted as follows:


n
string_1
...
string_n


Each dataset consists of n+1 lines. The first line of each dataset contains an integer that indicates n (1 \leq n \leq 500). The i-th line of the following n lines contains string_i, which consists of up to 10 English lowercase letters.

The end of the input is `0`, and this should not be processed.

Output

Print either `yes` or `no` in a line for each dataset, in the order of the input. If all words in the dataset can be considered to be ordered lexicographically, print `yes`. Otherwise, print `no`.

Example

Input

4
cba
cab
b
a
3
bca
ab
a
5
abc
acb
b
c
c
5
abc
acb
c
b
b
0


Output

yes
no
yes
no
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
typedef long long int ll;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000001
using namespace std;

enum RESULT{
	different,
	pre_shorter,
	pre_longer,
};

char table[26][26];

struct Info{
	char pre,current;
	RESULT result;
};

void strcpy(char* to,char* str){
	for(int i=0;str[i] != '\0';i++){
		to[i] = str[i];
		to[i+1] = '\0';
	}
}

Info compare(char pre[11],char current[11]){
	Info info;

	int pre_length,current_length;

	for(pre_length = 0;pre[pre_length] != '\0';pre_length++);
	for(current_length = 0; current[current_length] != '\0';current_length++);

	int index;

	for(index = 0;pre[index] == current[index] && index < pre_length && index < current_length ;index++);

	if(index == pre_length && index == current_length){
		info.result = pre_shorter;
	}else if(index != pre_length && index == current_length){
		info.result = pre_longer;
	}else if(index == pre_length && index != current_length){
		info.result = pre_shorter;
	}else{

		info.result = different;
		info.pre = pre[index];
		info.current = current[index];
	}

	return info;
}


int main(){

	int N;
	char pre[11],current[11];
	bool FLG;
	Info reason;

	while(true){
		scanf("%d",&N);
		if(N == 0)break;

		for(int i = 0; i < 26; i++){
			for(int k = 0; k < 26; k++){
				if(i != k){
					table[i][k] = '?';
				}else{
					table[i][k] = 'a'+ i;
 				}
			}
		}

		FLG = true;

		scanf("%s",pre);

		for(int i = 1; i < N; i++){
			scanf("%s",current);

			if(FLG){
				reason = compare(pre,current);
				if(reason.result == pre_longer){
					FLG = false;
				}else if(reason.result == different){

					if(table[reason.pre-'a'][reason.current-'a'] == '?'){

						for(int k = 0; k < 26; k++){
							if(table[k][reason.pre-'a'] == 'a'+k && table[k][reason.current-'a'] == reason.current){
								FLG = false;
								break;
							}
						}

						table[reason.pre-'a'][reason.current-'a'] = reason.pre;
						table[reason.current-'a'][reason.pre-'a'] = reason.pre;

						for(int me = 0;me < 26; me++){
							for(int you = 0; you < 26; you++){
								if(me != you && table[me][you] == 'a' + me){
									for(int other = 0; other < 26; other++){
										if(you != other && table[you][other] == 'a' + you){
											table[me][other] = 'a' + me;
											table[other][me] = 'a' + me;
										}
									}
								}
							}
						}

					}else if(table[reason.pre-'a'][reason.current-'a'] == reason.pre){
						//Do nothing
					}else{
						FLG = false;
					}
				}
				strcpy(pre,current);
			}
		}

		if(FLG)printf("yes\n");
		else{
			printf("no\n");
		}
	}

	return 0;
}