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 <string>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;

int n;
string str[501];
vector<int> G[27];
bool used[27];
int edge[27][27];
int num[27];
vector<int> vs;
void dfs(int v){
	used[v]=true;
	for(int i=0;i<G[v].size();i++){
		if(!used[G[v][i]])dfs(G[v][i]);
	}
	vs.push_back(v);
}

bool check(){
	for(int i=0;i<27;i++){
		G[i].clear();
	}
	memset(edge,0,sizeof(edge));
	for(int i=0;i<n;i++){
		for(int j=i+1;j<n;j++){
			int k;
			for(k=0;k<min(str[i].size(),str[j].size());k++){
				if(str[i][k]!=str[j][k]){
					if(edge[str[i][k]-'a'][str[j][k]-'a']==0){
						G[str[i][k]-'a'].push_back(str[j][k]-'a');
						edge[str[i][k]-'a'][str[j][k]-'a']=1;
					}
					break;
				}
			}
			if(k==min(str[i].size(),str[j].size()) && str[i].size()>str[j].size())return false;
		}
	}
	memset(used,0,sizeof(used));
	for(int i=0;i<=26;i++){
		if(!used[i])dfs(i);
	}
	for(int i=0;i<vs.size();i++){
		num[vs[i]]=i;
	}
	for(int i=0;i<26;i++){
		for(int j=0;j<G[i].size();j++){
			if(num[i]<num[G[i][j]])return false;
		}
	}
	return true;
}

int main(void){
	while(1){
		scanf("%d",&n);
		if(n==0)break;
		for(int i=0;i<n;i++){
			cin >> str[i];
		}
		printf("%s\n",check()?"yes":"no");
	}
	return 0;
}