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 <cstdio>
#include <cstring>
char ss[510][20];
bool map[30][30];
bool v[30];
int main(){
	int n;
	while (scanf("%d", &n) && n){
		for (int i = 0; i < n; i ++){
			scanf("%s", ss[i]);
			for (int j = 0; ss[i][j]; j ++){
				ss[i][j] -= 'a' - 1;
			}
		}
		bool mark = false;
		memset(map, 0, sizeof(map));
		for (int i = 0; i < n; i ++)
			for (int j = i + 1; j < n; j ++){
				for (int k = 0; ss[i][k] || ss[j][k]; k ++)
					if (ss[i][k] != ss[j][k]){
						if (ss[i][k] && !ss[j][k])
							mark = true;
						if (ss[i][k] && ss[j][k])
							map[ss[i][k]][ss[j][k]] = 1;
						break;
					}
			}
		for (int k = 1; k <= 26; k ++)
			for (int i = 1; i <= 26; i ++)
				for (int j = 1; j <= 26; j ++)
					if (map[i][k] && map[k][j])
						map[i][j] = true;
		for (int i = 1; i <= 26; i ++)
			for (int j = 1; j <= 26; j ++)
				if (map[i][j] && map[j][i])
					mark = true;
		if (mark)
			puts("no");
		else
			puts("yes");
	}
	return 0;
}