Codeforces Round #619 (Div. 2) A. Three Strings

You are given three strings aa , bb and cc of the same length nn . The strings consist of lowercase English letters only. The ii -th letter of aa is aiai , the ii -th letter of bb is bibi , the ii -th letter of cc is cici .

For every ii (1≤i≤n1≤i≤n ) you must swap (i.e. exchange) cici with either aiai or bibi . So in total you'll perform exactly nn swap operations, each of them either ci↔aici↔ai or ci↔bici↔bi (ii iterates over all integers between 11 and nn , inclusive).

For example, if aa is "code", bb is "true", and cc is "help", you can make cc equal to "crue" taking the 11 -st and the 44 -th letters from aa and the others from bb . In this way aa becomes "hodp" and bb becomes "tele".

Is it possible that after these swaps the string aa becomes exactly the same as the string bb ?

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1001≤t≤100 )  — the number of test cases. The description of the test cases follows.

The first line of each test case contains a string of lowercase English letters aa .

The second line of each test case contains a string of lowercase English letters bb .

The third line of each test case contains a string of lowercase English letters cc .

It is guaranteed that in each test case these three strings are non-empty and have the same length, which is not exceeding 100100 .

Output

Print tt lines with answers for all test cases. For each test case:

If it is possible to make string aa equal to string bb print "YES" (without quotes), otherwise print "NO" (without quotes).

You can print either lowercase or uppercase letters in the answers.

Example

Input

4
aaa
bbb
ccc
abc
bca
bca
aabb
bbaa
baba
imi
mii
iim

Output

NO
YES
YES
NO
水题。最终目的是要求a和b一样，这样的话，因为ai或者bi其中之一可以与ci交换，所以如果ai==ci,那么拿bi与ci交换，之后可以得到ai==bi;同理bi==ci，可以拿ai与ci交换，遍历一遍看看能不能满足即可。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        char a[];
        char b[];
        char c[];
        scanf("%s",a);
        scanf("%s",b);
        scanf("%s",c);
        int i;
        int cnt=;
        for(i=;i<strlen(a);i++)
        {
            if(a[i]==c[i]||b[i]==c[i])cnt++;
        }
        if(cnt==strlen(a))
        {
            cout<<"YES"<<endl;
            continue;
        }
        cout<<"NO"<<endl;
    }
}