Suppose you have an integer array $a_1, a_2, \dots, a_n$.

Let $\operatorname{lsl}(i)$ be the number of indices $j$ ($1 \le j \lt i$) such that $a_j \lt a_i$.

Analogically, let $\operatorname{grr}(i)$ be the number of indices $j$ ($i \lt j \le n$) such that $a_j \gt a_i$.

Let's name position $i$ good in the array $a$ if $\operatorname{lsl}(i) \lt \operatorname{grr}(i)$.

Finally, let's define a function $f$ on array $a$ $f(a)$ as the sum of all $a_i$ such that $i$ is good in $a$.

Given two integers $n$ and $k$, find the sum of $f(a)$ over all arrays $a$ of size $n$ such that $1 \leq a_i \leq k$ for all $1 \leq i \leq n$ modulo $998\,244\,353$.

Input

The first and only line contains two integers $n$ and $k$ ($1 \leq n \leq 50$; $2 \leq k \lt 998\,244\,353$).

Output

Output a single integer — the sum of $f$ over all arrays $a$ of size $n$ modulo $998\,244\,353$.

Note

In the first test case:

$f([1,1,1]) = 0$$f([2,2,3]) = 2 + 2 = 4$$f([1,1,2]) = 1 + 1 = 2$$f([2,3,1]) = 2$$f([1,1,3]) = 1 + 1 = 2$$f([2,3,2]) = 2$$f([1,2,1]) = 1$$f([2,3,3]) = 2$$f([1,2,2]) = 1$$f([3,1,1]) = 0$$f([1,2,3]) = 1$$f([3,1,2]) = 1$$f([1,3,1]) = 1$$f([3,1,3]) = 1$$f([1,3,2]) = 1$$f([3,2,1]) = 0$$f([1,3,3]) = 1$$f([3,2,2]) = 0$$f([2,1,1]) = 0$$f([3,2,3]) = 2$$f([2,1,2]) = 1$$f([3,3,1]) = 0$$f([2,1,3]) = 2 + 1 = 3$$f([3,3,2]) = 0$$f([2,2,1]) = 0$$f([3,3,3]) = 0$$f([2,2,2]) = 0$

Adding up all of these values, we get $28$ as the answer.