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 <iostream>
#include <algorithm>
#include <iomanip>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using u32 = uint32_t;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;

int solve(int n){
    int M = 26;
    vector<int> deg(M);
    vector<set<int>> G(M);
    vector<string> s(n);
    for (int i = 0; i < n; ++i) {
        cin >> s[i];
    }
    for (int i = 0; i < n-1; ++i) {
        int cur = 0;
        while(s[i][cur] == s[i+1][cur] &&
              cur < min(s[i].size(), s[i+1].size())) cur++;
        if(cur == min(s[i].size(), s[i+1].size())){
            if(s[i].size() <= s[i+1].size()) continue;
            else return 0;
        }
        G[s[i][cur]-'a'].emplace(s[i+1][cur]-'a');
        if(G[s[i+1][cur]-'a'].count(s[i][cur]-'a')) return 0;
    }
    for (int i = 0; i < M; ++i) {
        for (auto &&j : G[i]) {
            deg[j]++;
        }
    }
    vector<int> visited(M);
    stack<int> Q;
    for (int i = 0; i < M; ++i) {
        if(!deg[i]) Q.emplace(i);
    }
    while(!Q.empty()){
        int i = Q.top(); Q.pop();
        visited[i] = 1;
        for (auto &&j : G[i]) {
            deg[j]--;
            if(!deg[j]){
                Q.emplace(j);
            }
        }
        G[i].clear();
    }
    for (int i = 0; i < M; ++i) {
        if(!visited[i]) return 0;
    }
    return 1;
}

int main() {
    int n;
    while(cin >> n, n){
        puts(solve(n) ? "yes" : "no");
    }

    return 0;
}
