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>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<29;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-11;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

class Graph{
	public:
	int n;
	vvi g,rg;
	vi scc,used;
	stack<int> stc;
	void Order(int v){
		used[v]=1;
		for(auto u:g[v]) if(!used[u]) Order(u);
		stc.push(v);
	}
	void DFS(int v,int t){
		used[v]=1;
		for(auto u:rg[v]) if(!used[u]) DFS(u,t);
		scc[v]=t;
	}
	int SCC(){
		scc=used=vi(n);
		rg=vvi(n);
		for(int i=0;i<n;i++){
			if(!used[i]) Order(i);
			for(auto v:g[i]) rg[v].push_back(i);
		}
		int t=0;
		used=vi(n);
		while(!stc.empty()){
			int v=stc.top();
			stc.pop();
			if(!used[v]) DFS(v,t++);
		}
		return t;
	}
	Graph(int v){
		n=v;
		g=vvi(v);
	}
	void add_edge(int s,int t){
		g[s].push_back(t);
	}
};

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	while(1){
		int n;
		cin>>n;
		if(n==0) break;
		vs a(n);
		for(int i=0;i<n;i++) cin>>a[i];
		bool B=0;
		Graph g(26);
		for(int i=1;i<n;i++){
			int s=a[i-1].size(),s1=a[i].size();
			bool flag=0;
			for(int j=0;j<min(s,s1);j++) if(a[i-1][j]!=a[i][j]){
				g.add_edge(a[i-1][j]-'a',a[i][j]-'a');
				flag=1;
				break;
			}
			if(!flag&&s>s1) B=1;
		}
		if(B){
			cout<<"no"<<endl;
			continue;
		}
		if(g.SCC()<26) cout<<"no"<<endl;
		else cout<<"yes"<<endl;
	}
}
