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 rep(i, a, n) for(int i = a; i < n; i++)
#define int long long
using namespace std;
typedef pair<int, int> P;

bool G[26][26];
bool f[26];
bool ans;

void dfs(int now, int root){
    // cout << now << endl;
    if(f[now]){
        if(now == root) ans = false;
        return;
    }
    f[now] = true;
    rep(i, 0, 26){
        if(G[now][i]) dfs(i, root);
    }
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    while(cin >> n, n){
        vector<string> s(n);
        ans = true;
        rep(i, 0, 26) rep(j, 0, 26) G[i][j] = false;
        rep(i, 0, n){
            cin >> s[i];
        } 
        rep(i, 0, n - 1){
            int lm = min(s[i].size(), s[i + 1].size());
            bool same = true;
            rep(j, 0, lm){
                if(s[i][j] == s[i + 1][j]) continue;
                else{
                    same = false;
                    G[s[i][j] - 'a'][s[i + 1][j] - 'a'] = true;
                    break;
                }
            }
            if(same){
                if(s[i].size() > s[i + 1].size()){
                    ans = false;
                    break;
                }
            }
        }
        rep(i, 0, 26){
            memset(f, false, sizeof(f));
            dfs(i, i);
        }
        if(ans) cout << "yes" << endl;
        else cout << "no" << endl;
    }
}
