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

int main()
{
	ios::sync_with_stdio(false), cin.tie(0);
	int n;
	while (cin >> n, n) {
		vector<string> s(n);
		for (int i = 0; i < n; i++) {
			cin >> s[i];
		}
		bool ok = true;
		vector<vector<int>> d(26, vector<int>(26, 1e9));
		for (int i = 1; i < n; i++) {
			int l1 = s[i - 1].size(), l2 = s[i].size();
			int p = 0;
			while (p < l1 && p < l2 && s[i - 1][p] == s[i][p]) ++p;
			if (p < l1 && p < l2) {
				d[s[i - 1][p] - 'a'][s[i][p] - 'a'] = -1;
			}
			else if (p < l1) {
				ok = false;
			}
		}
		for (int k = 0; k < 26; k++) {
			for (int i = 0; i < 26; i++) {
				for (int j = 0; j < 26; j++) {
					d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
				}
			}
		}
		for (int i = 0; i < 26; i++) {
			if (d[i][i] < 0) {
				ok = false;
			}
		}
		puts(ok ? "yes" : "no");
	}
	return 0;
}

