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;

#define fi first
#define se second
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb(s) push_back(s)
#define mp(a,b) make_pair(a,b)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define maxch(x,y) x=max(x,y)
#define minch(x,y) x=min(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt(x) bitset<32>(x).count()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<P, int> PPI;
typedef pair<ll, ll> PL;
typedef pair<P, ll> PPL;

#define INF INT_MAX/3

#define MAX_N 1000

#define next nxt

int n;
string s[555];
vector<int> next[33];
bool vis[33];
int sch;

bool dfs(int v){
	vis[v]=true;
	bool res=true;
	rep(i,next[v].size()){
		int nv=next[v][i];
		if(nv==sch)res=false;
		if(vis[nv])continue;
		if(!dfs(nv))res=false;
	}
	return res;
}

int main(){
	cin.sync_with_stdio(false);
	while(1){
		cin>>n;
		if(n==0)break;
		rep(i,n)cin>>s[i];
		rep(i,33)next[i].clear();
		bool ok=true;
		rep(i,n-1){
			int j=0;
			while(j<min(s[i].length(),s[i+1].length())&&s[i][j]==s[i+1][j])j++;
			if(j==s[i].length())continue;
			if(j==s[i+1].length()){
				ok=false;
				break;
			}
			if(!exist(next[s[i][j]-'a'],s[i+1][j]-'a'))next[s[i][j]-'a'].pb(s[i+1][j]-'a');
		}
		rep(i,33){
			memset(vis,0,sizeof(vis));
			sch=i;
			if(!dfs(i)){
				ok=false;
				//dbg(i);
			}
		}
		if(!ok)cout<<"no"<<endl;
		else cout<<"yes"<<endl;
	}
	return 0;
}