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;
set<int> G[26];
typedef set<int>::iterator it;
int n;
string s[505];
bool flag;
bool vis[26];
int in[26];
int main()
{
	while(~scanf("%d",&n)&&n)
	{
		for(int i=0;i<n;i++)
			cin>>s[i];
		for(int i=0;i<26;i++)
			G[i].clear(),vis[i] = 0,in[i] = 0;
		flag = true;
		for(int x=1;x<n;x++)
			for(int y=0;y<x;y++)
		    {
			    int len1 = s[y].size();
			    int len2 = s[x].size();
			    bool flag1 = true;
			    for(int i=0;i<min(len1,len2);i++)
				    if(s[y][i]!=s[x][i])
			        {
			    	    flag1 = false;
			    	    G[s[y][i]-'a'].insert(s[x][i]-'a');
			    	    break;
			        }
			    if(flag1&&len1>len2)
				    flag = false;
		    }
		int cnt = 0;
		for(int i=0;i<26;i++)
			for(it p = G[i].begin();p!=G[i].end();p++)
			    in[*p]++,cnt++;
		while(1)
		{
			bool flag1 = true;
			for(int i=0;i<26;i++)
				if(in[i]==0&&!vis[i])
				{
					flag1 = false;
					vis[i] = 1;
				    for(it p = G[i].begin();p != G[i].end();p++)
			        {
			        	int v = *p;
			        	cnt--;
			        	in[v]--;
			        }
				}
			if(flag1) break;
		}
		if(cnt!=0) flag = false;
		puts(flag?"yes":"no");
	}
	return 0;
}