Restoration of string （构图+思维）

A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.

You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).

A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.

The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.

String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of strings in the set.

Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.

The total length of the strings doesn't exceed 10⁵.

Output

Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.

Examples

Input

4
mail
ai
lru
cf

Output

cfmailru

Input

3
kek
preceq
cheburek

Output

NO

Note

One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.

题意：构造一个字符串str要求所给的字符串是该串的字串，在str的所有字串中，出现的频率的是最高的，如果有多个答案，输出

输出字典序，最小的一个。

思路：构成的字符串所有字母只能出现一次，首先将每个字符串构成一条有方向的链，表示前一个字母可以到达后一个字母

，然后遍历所有的字符串链，将这些链构成的单词连起来就是所求字符串

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
using namespace std;
const int N =  1e5+10;
string a[N],ans;
int in[30],out[30];
int vis[30],p[30];
int d[30][30];
int n;
int Dfs(int now){
	vis[now]=1;
	ans=ans+(char)(now+'a');
	for(int i=0;i<26;i++){
		if(d[now][i]){
			if(vis[i]) return 0; //被访问过，说明其他链有这个字母，字母出现多次 
			return Dfs(i);
		}
	}
	return 1;
}
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		cin>>a[i];
		int len=a[i].length();
		for(int j=0;j<len-1;j++){
			int x=a[i][j]-'a';
			int y=a[i][j+1]-'a';
			d[x][y]=1;//建一条从x->y的有向边 
		}
		if(len==1) p[a[i][0]-'a']=1; //单个点 
	}
	
	for(int i=0;i<26;i++)
	   if(d[i][i]){ //判断自环,说明改字母出现了多次 
	   	 printf("NO\n");
	   	 return 0;
	   }
	
	for(int i=0;i<26;i++)
	for(int j=0;j<26;j++)
	   if(d[i][j]) in[j]++,out[i]++;
	
	for(int i=0;i<26;i++)
	   if(in[i]>1||out[i]>1){
	   	  printf("NO\n");
	   	  return 0;
	   }
	
	ans="";
	for(int i=0;i<26;i++){ //字母从小到大加，保证新字符串的字典序最小 
		if(out[i]!=0&&in[i]==0){ //从某条字符串链，起点开始，入度为0，出度为1 
			if(!Dfs(i)){
				printf("NO\n");
				return 0;
			}
		}
		if(p[i]&&in[i]==0&&out[i]==0) ans=ans+(char)(i+'a'); //只有一个字母 
	}
	
	for(int i=0;i<26;i++){
		if(out[i]!=0||in[i]!=0){
			if(!vis[i]){ //在DFS中没有被标记，说明构成了环 
				printf("NO\n");
				return 0;
			}
		}
	}
	cout<<ans<<endl;   
	return 0;
}

Restoration of string （构图+思维）

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