CF1650A.Deletions of Two Adjacent Letters

The string $s$ is given, the string length is odd number. The string consists of lowercase letters of the Latin alphabet.

As long as the string length is greater than $1$ , the following operation can be performed on it: select any two adjacent letters in the string $s$ and delete them from the string. For example, from the string "lemma" in one operation, you can get any of the four strings: "mma", "lma", "lea" or "lem" In particular, in one operation, the length of the string reduces by $2$ .

Formally, let the string $s$ have the form $s=s_1s_2 \dots s_n$ ( $n>1$ ). During one operation, you choose an arbitrary index $i$ ( $1 \le i < n$ ) and replace $s=s_1s_2 \dots s_{i-1}s_{i+2} \dots s_n$ .

For the given string $s$ and the letter $c$ , determine whether it is possible to make such a sequence of operations that in the end the equality $s=c$ will be true? In other words, is there such a sequence of operations that the process will end with a string of length $1$ , which consists of the letter $c$ ?

The first line of input data contains an integer $t$ ( $1 \le t \le 10^3$ ) — the number of input test cases.

The descriptions of the $t$ cases follow. Each test case is represented by two lines:

• string $s$ , which has an odd length from $1$ to $49$ inclusive and consists of lowercase letters of the Latin alphabet;
• is a string containing one letter $c$ , where $c$ is a lowercase letter of the Latin alphabet.

For each test case in a separate line output:

• YES, if the string $s$ can be converted so that $s=c$ is true;
• NO otherwise.

You can output YES and NO in any case (for example, the strings yEs, yes, Yes and YES will be recognized as a positive response).

In the first test case, $s$ ="abcde". You need to get $s$ ="c". For the first operation, delete the first two letters, we get $s$ ="cde". In the second operation, we delete the last two letters, so we get the expected value of $s$ ="c".

In the third test case, $s$ ="x", it is required to get $s$ ="y". Obviously, this cannot be done.