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
// 文字に適切な順序を定めて、入力が辞書順になるようにできるか
// 適当に辺を貼って閉路見つける

#include <bits/stdc++.h>
using namespace std;

int n;
string s[500];
bool f[500][500];
bool g[27][27];

int main()
{
	while (cin >> n, n){
		memset(f, 0, sizeof(f));
		memset(g, 0, sizeof(g));
		for (int i = 1; i < 27; i++){
			g[0][i] = true;
		}

		for (int i = 0; i < n; i++){
			cin >> s[i];
			s[i] += 'a' - 1;
		}
		
		for (int i = 0; i <= 10; i++){
			for (int j = 0; j < n; j++){
				for (int k = j + 1; k < n; k++){
					if (f[j][k]) continue;
					if (i >= min(s[j].size(), s[k].size())) continue;
					if (s[j][i] != s[k][i]){
						g[s[j][i] - 'a' + 1][s[k][i] - 'a' + 1] = true;
						f[j][k] = true;
					}
				}
			}
		}

		for (int k = 0; k < 27; k++){
			for (int i = 0; i < 27; i++){
				for (int j = 0; j < 27; j++){
					g[i][j] |= g[i][k] && g[k][j];
				}
			}
		}
		
		bool res = true;
		for (int i = 0; i < 27; i++){
			if (g[i][i]) res = false;
		}
		
		cout << (res ? "yes" : "no") << endl;
	}
}