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The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.
Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!
Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.
The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.
The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.
For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.
You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).
Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case. |