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 F first
#define S second
#define pi M_PI
#define R cin>>
#define Z class
#define ll long long
#define ln cout<<'\n'
#define in(a) insert(a)
#define pb(a) push_back(a)
#define pd(a) printf("%.10f\n",a)
#define mem(a) memset(a,0,sizeof(a))
#define all(c) (c).begin(),(c).end()
#define iter(c) __typeof((c).begin())
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define REP(i,m,n) for(int i=(int)(m);i<(int)(n);i++)
#define rep(i,n) REP(i,0,n)
#define tr(it,c) for(iter(c) it=(c).begin();it!=(c).end();it++)
template<Z A>void pr(A a){cout<<a;ln;}
template<Z A,Z B>void pr(A a,B b){cout<<a<<' ';pr(b);}
template<Z A,Z B,Z C>void pr(A a,B b,C c){cout<<a<<' ';pr(b,c);}
template<Z A,Z B,Z C,Z D>void pr(A a,B b,C c,D d){cout<<a<<' ';pr(b,c,d);}
template<Z A>void PR(A a,ll n){rep(i,n){if(i)cout<<' ';cout<<a[i];}ln;}
ll check(ll n,ll m,ll x,ll y){return x>=0&&x<n&&y>=0&&y<m;}
const ll MAX=1000000007,MAXL=1LL<<61,dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
typedef pair<int,int> P;

#define N 100
vector<int> G[N],rG[N],g;
bool u[N];
int n,cmp[N];
  
void add_edge(int x,int y){G[x].push_back(y);rG[y].push_back(x);}
void dfs(int x){u[x]=1;rep(i,G[x].size())if(!u[G[x][i]])dfs(G[x][i]);g.push_back(x);} 
void rdfs(int x,int k){u[x]=1;cmp[x]=k;rep(i,rG[x].size())if(!u[rG[x][i]])rdfs(rG[x][i],k);}
int scc() {
  mem(u);g.clear();rep(i,n)if(!u[i])dfs(i);mem(u);
  int k=0;rrep(i,g.size())if(!u[g[i]])rdfs(g[i],k++);return k;
}

void Main() {
  n=26;
  int m;
  while(R m && m) {
    rep(i,n) G[i].clear(),rG[i].clear();
    set<int> se;
    string s[m];
    rep(i,m) R s[i];
    rep(i,m-1) {
      bool f=1;
      rep(j,min(s[i].size(),s[i+1].size())) {
        if(s[i][j]!=s[i+1][j]) {
          add_edge(s[i][j]-'a',s[i+1][j]-'a');
          f=0;
          break;
        }
      }
      if(f&&s[i].size()>s[i+1].size()) {
        pr("no");
        goto next;
      }
    }
    scc();
    rep(i,26) se.in(cmp[i]);
    if(se.size()==n) pr("yes");
    else pr("no");
  next:;
  }

}

int main(){ios::sync_with_stdio(0);cin.tie(0);Main();return 0;}
