找子串

B. Substrings Sort

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given $n$ strings. Each string consists of lowercase English letters. Rearrange (reorder) the given strings in such a way that for every string, all strings that are placed before it are its substrings.

String $a$ is a substring of string $b$ if it is possible to choose several consecutive letters in $b$ in such a way that they form $a$. For example, string "for" is contained as a substring in strings "codeforces", "for" and "therefore", but is not contained as a substring in strings "four", "fofo" and "rof".

Input

The first line contains an integer $n$ ($1\le n\le 100$) — the number of strings.

The next $n$ lines contain the given strings. The number of letters in each string is from $1$ to $100$, inclusive. Each string consists of lowercase English letters.

Some strings might be equal.

Output

If it is impossible to reorder $n$ given strings in required order, print "NO" (without quotes).

Otherwise print "YES" (without quotes) and $n$ given strings in required order.

Examples

input

Copy

5
a
aba
abacaba
ba
aba

output

Copy

YES
a
ba
aba
aba
abacaba

input

Copy

5
a
abacaba
ba
aba
abab

output

Copy

NO

input

Copy

3
qwerty
qwerty
qwerty

output

Copy

YES
qwerty
qwerty
qwerty

Note

In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".

题意:先按长度排列字符串,然后从小到大前一项是后一项的子串,直到最后一个

#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
string str[110];
int n;
bool cmp(string a,string b)
{
    return a.size()<b.size();
}
int main()
{
    bool flag=true;
    cin>>n;
    for(int i=0;i<n;i++)
     cin>>str[i];
    sort(str,str+n,cmp);
    for(int i=0;i<n-1;i++)
        if(str[i+1].find(str[i])==string::npos)//找不到子串
           flag=false;
    if(!flag)  puts("NO");
    else
    {
        puts("YES");
        for(int i=0;i<n;i++)
            cout<<str[i]<<endl;
    }

    return 0;
}