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;

const int M = 26;

char g[M + 1][M + 1]; // g[i][j] char[i] > char[j]
int visit[M + 1];

bool dfs(int i) {
	visit[i] = 1;
	for(int j = 0;j <= M;++j){
		if(g[i][j]) {
			if(visit[j] == 2) continue;
			if(visit[j] == 1) return false;
			if(!dfs(j)) return false;
		}
	}
	visit[i] = 2;
	return true;
}

void ordering(string &a, string &b){
	int len = max(a.length(), b.length());
	for(int i = 0;i < len;++i){
		if(a.length() == i){
			break;
		}
		else if(b.length() == i){
			g[M][a[i] - 'a'] = 1;
			break;
		}
		else if(a[i] != b[i]){
			g[b[i] - 'a'][a[i] - 'a'] = 1;
			break;
		}
	}
	return;
}

int main() {
	int n;
	string s;
	bool judge;
	while(cin >> n, n) {
		vector<string> seq;
		set<char> m;
		memset(g, 0, sizeof(g));
		memset(visit, 0, sizeof(visit));
		judge = true;
		for(int i = 0;i < n;++i){
			cin >> s;
			seq.emplace_back(s);
		}
		g[seq[0][0] - 'a'][M] = 1;
		for(int i = 1;i < n;++i){
			ordering(seq[i - 1], seq[i]);
		}
		for(int i = 0;i <= M;++i){
			judge = judge && dfs(i);
		}
		if(judge){
			cout << "yes" << endl;
		}
		else{
			cout << "no" << endl;
		}
	}
	return 0;
}

