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;

vector<int> G[30];
bool g[30];

int d[30][30];
const int INF = 1e5;

/*bool dfs(int now, int pre){
	if(g[now] == true) return false;
	bool res = true;
	//cout << now << endl;
	g[now] = true;
	set<int> st;
	for(int i = 0; i < G[now].size(); i++){
		int next = G[now][i];
		//if(next == pre) continue;
		if(st.count(next)) continue;
		st.insert(next);
		res &= dfs(next, now);
	}
	return res;
}*/

void wf(){
	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]);
			}
		}
	}
	return;
}

int main(){
	int n;
	while(1){
		for(int i = 0; i < 26; i++){
			G[i].clear();
			for(int j = 0; j < 26; j++) d[i][j] = INF;
			d[i][i] = 0;
		}
		cin >> n;
		if(n == 0) break;
		vector<string> s(n);
		for(int i = 0; i < n; i++){
			cin >> s[i];
		}
		bool f = true;
		for(int i = 0; i < n - 1; i++){
			for(int j = 0; j < s[i].size(); j++){
				if(s[i + 1].size() == j){
					f = false; break;
				}
				if(s[i][j] != s[i + 	1][j]){
					G[s[i][j] - 'a'].push_back(s[i + 1][j] - 'a');
					//cout << s[i][j] << "->" << s[i + 1][j] << endl;
					break;
				}
			}
		}
		for(int i = 0; i < 26; i++){
			for(int j = 0; j < G[i].size(); j++){
				d[i][G[i][j]] = -1;
			}
		}
		wf();
		//int e[26][26] = {};
		//for(int i = 0; i < 26; i++) for(int j = 0; j < 26; j++) e[i][j] = d[i][j];
		//wf();
		for(int i = 0; i < 26; i++){
			//cout << d[i][i] << endl;
			if(d[i][i] != 0) f = false;
			//for(int j = 0; j < 26; j++){
			//	if(d[i][j] != e[i][j]) f = false;
			//}
		}
		/*for(int j = 0; j < 26; j++){
		//cout << j << " " << f << endl;
			for(int i = 0; i < 26; i++) g[i] = false;
			f &=  dfs(j, -1);
		}*/
		if(f) cout << "yes" << endl;	
		else cout << "no" << endl;
	}
}