Problem Description

Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string s of length n. He wants to find three nonoverlapping substrings s[l1..r1], s[l2..r2], s[l3..r3] that:

1. 1≤l1≤r1<l2≤r2<l3≤r3≤n
2. The concatenation of s[l1..r1], s[l2..r2], s[l3..r3] is "anniversary".

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤100), indicating the number of test cases. For each test case:

There's a line containing a string s (1≤|s|≤100) consisting of lowercase English letters.

 

Output

For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).

 

Sample Input

 

Sample Output

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#define  N 150

char s1[] = "anniversary";

char ch[N];

char s[N][][];

void Init()

{

    int j, z, w=;

    for(j=; j<=; j++)

    for(z=; z<=-j; z++)

    {

        strncpy(s[w][], s1, j);

        strncpy(s[w][], s1+j, z);

        strcpy(s[w++][], s1+j+z);

    }

}

int main()

{

   int t;

   scanf("%d", &t);

   Init();

   while(t--)

   {

       char ch[N];

       int i, a, b, len1, len2, flag=;

       memset(ch, , sizeof(ch));

       scanf("%s", ch);

       for(i=; i<; i++)

       {

           if(strstr(ch, s[i][]))

           {

               a = strstr(ch, s[i][])-ch;

               len1 = strlen(s[i][]);

               a = a + len1;

               if(strstr(ch+a, s[i][]))

               {

                   b = strstr(ch+a, s[i][])-(ch+a); ///就在这， 我在判断的时候还把字符串向后移了 a 位， 但在                   ///计算的时候并没有用，这是最大的失误，难怪自己思路对了但一直提交错误

                   len2 = strlen(s[i][]);

                   b = a + b + len2;

                   if(strstr(ch+b, s[i][]))

                   {

                       flag = ;

                       break;

                   }

               }

           }

       }

       if(flag)  printf("YES\n");

       else  printf("NO\n");

   }

    return ;

}