B. Substrings Sort

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given nn 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 aa is a substring of string bb if it is possible to choose several consecutive letters in bb in such a way that they form aa. 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 nn (1≤n≤1001≤n≤100) — the number of strings.

The next nn lines contain the given strings. The number of letters in each string is from 11 to 100100, inclusive. Each string consists of lowercase English letters.

Some strings might be equal.

Output

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

Otherwise print "YES" (without quotes) and nn 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：给你一堆字符串，能不能把它们排序，排成从上到下，上面一个是下面一个的子字符串的格式，能就按格式全部输出，不能就输出“NO”.

#include<bits/stdc++.h>
using namespace std;
bool cmp(string a,string b){
	if(a.length()==b.length()){
		return a<b;
	}return a.length()<b.length();   //先把所有字符串排序 
}                                    //按字符串长度，字典序排序 
string a[200];
int main(){
	int m;
	cin>>m;
	for(int i=0;i<m;i++){
		cin>>a[i];
	}
	sort(a,a+m,cmp);
	
	bool ff=true;
	for(int i=0;i<m-1;i++){                 //遍历所有 
	    bool f=true;
		int t=a[i].length();
		string temp=a[i+1];
		int tt=a[i+1].length();
		for(int j=0;j<=tt-t;j++){           
			if(temp.substr(j,t)==a[i]){    //两个两个检查，假如其中有一组(两个相邻)的，不合要求，标记 
				f=false;                  
			}
		}
		if(f){
			ff=false;         //利用标记退出； 
			break;
		}
	}
	if(!ff){
		cout<<"NO"<<endl;
	}
	else {
		cout<<"YES"<<endl;
		for(int i=0;i<m;i++){
			cout<<a[i]<<endl;
		}
	}
	return 0;
}