Codeforces Round #486 (Div. 3)-B. Substrings Sort

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".

被HACK掉了。。。只能说数据太水了。。。发现自己的程序问题还这么大，还是太菜了。

本题题意就是是否可以把所给的N个序列摆放成上面的序列是下面序列的子序列

开始想了半天。。。怎么找啊。。。这™序列这么多，要我一个一个比？后来发现，你只需要按照从小到大的序列长度排序就好了啊。。。然后从下往上，两个判断是否上面的那么个序列是下面序列的子序列。o(n)操作嘛，小case.

被HACK掉原因就是我匹配那里写错了

正确写匹配姿势:

  for (int i=; i<=word[p].len-word[p-].len; i++)//每个头位置的指针
     {
         flag=;
         pi=i;//检查在I为头的情况下的是否匹配的指针
         for(int j=; j<word[p-].len;j++)
         {
             if(word[p].sub[pi]!=word[p-].sub[j])
             {
                 flag=;//如果有不同
                 break;
             }
             else
             {
                 pi++;//继续匹配
                 continue;
             }
         }
         if (flag==)break;
     }

emmmmm最后AC代码

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct Node
{
    char sub[];
    int len;
} word[];
bool cmp(Node x,Node y)
{
    return x.len<y.len;//按照长度排序
}
int pp(int p)
{
    int flag;
    int pi;
    for (int i=; i<=word[p].len-word[p-].len; i++)
    {
        flag=;
        pi=i;
        for(int j=; j<word[p-].len;j++)
        {
            if(word[p].sub[pi]!=word[p-].sub[j])
            {
                flag=;
                break;
            }
            else
            {
                pi++;
                continue;
            }
        }
        if (flag==)break;
    }
    if (flag==)return ;
    else return ;
}
int main()
{
    int n;
    int lens;
    int flag;
    while (~scanf("%d",&n))
    {
        flag=;
        for (int i=; i<n; i++)
        {
            scanf("%s",word[i].sub);
            lens=strlen(word[i].sub);
            word[i].len=lens;
        }
        sort(word,word+n,cmp);
        for (int i=n-; i>; i--)
        {
            if(pp(i)!=){
            flag=;
            break;
            }
        }
        if(flag==)printf("NO\n");
        else
        {
            printf("YES\n");
            for (int i=; i<n; i++)
            {
                printf("%s\n",word[i].sub);
            }
        }
    }
    return ;
}