2016 Al-Baath University Training Camp Contest-1 F

Description

Zaid has two words, a of length between 4 and 1000 and b of length 4 exactly. The word a is 'good' if it has a substring which is equal tob. However, a is 'almost good' if by inserting a single letter inside of it, it would become 'good'. For example, if a = 'start' and b = 'tear': bis not found inside of a, so it is not 'good', but if we inserted the letter 'e' inside of a, it will become 'good' ('steart'), so a is 'almost good' in this case. Your task is to determine whether the word a is 'good' or 'almost good' or neither.

Input

The input consists of several test cases. The first line of the input contains a single integer T, the number of the test cases. Each of the following T lines represents a test case and contains two space separated strings a and b, each of them consists of lower case English letters. It is guaranteed that the length of a is between 4 and 1000, and the length of b is exactly 4.

Output

For each test case, you should output one line: if a is 'good' print 'good', if a is 'almost good' print 'almost good', otherwise print 'none'.

Example

input

4
smart mark
start tear
abracadabra crab
testyourcode your

output

almost good
almost good
none
good

Note

A substring of string s is another string t that occurs in s. Let's say we have a string s = "abcdefg" Possible valid substrings: "a","b","d","g","cde","abcdefg". Possible invalid substrings: "k","ac","bcef","dh".

题意：两个字符串，如果b的长度为4且为a的子串的话，输出good，如果需a要加一个字符才能符合要求的话，输出almost good，否则输出none

题解：用find就行，b可以分解为三个字符组成的字符串，再判断就行

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    string s1,s2;
    cin>>n;
    while(n--)
    {
        cin>>s1>>s2;
        if(s1.find(s2)!=-1)
        {
            cout<<"good"<<endl;
        }
        else
        {
            int flag=0;
            for(int i=0; i<s2.length(); i++)
            {
                string s3="";
                for(int j=0; j<s2.length(); j++)
                {
                    if(i!=j)
                    {
                        s3+=s2[j];
                    }
                }
                if(s1.find(s3)!=-1)
                {
                    flag=1;
                }
               // cout<<s1.find(s3)<<endl;
            }
            if(flag)
            {
                cout<<"almost good"<<endl;
            }
            else
            {
                cout<<"none"<<endl;
            }
        }
    }
    return 0;
}