B'Vika likes playing with bracket sequences. Today she wants to create a new bracket sequence using the following algorithm. Initially, Vika 's sequence is an empty string, and then she will repeat the following actions n times: A bracket sequence is called regular if it is possible to obtain a correct arithmetic expression by inserting characters '+ ' and '1 ' into it. For example, sequences "(())()", "()", and "(()(()))" are regular, while ")(", "(()", and "(()))(" are not. Vika wants to know the probability that her bracket sequence will be a regular one at the end. Help her and find this probability modulo 998 ,244 ,353 (see Output section). The only line contains two integers n and q ( 1 <= n <= 500 ; 0 <= q <= 10^4 ). Here n is equal to the number of bracket insertion operations, and the probability that Vika chooses string "()" on every step of the algorithm is equal to p = q cdot 10^{-4} . Print the probability that Vika 's final bracket sequence will be regular, modulo 998 ,244 ,353 . Formally, let M = 998 ,244 ,353 . It can be shown that the answer can be expressed as an irreducible fraction frac{p}{q} , where p and q are integers and q not equiv 0 pmod{M} . Output the integer equal to p cdot q^{-1} bmod M . In other words, output such an integer x that 0 <= x < M and x cdot q equiv p pmod{M} . In the first example, Vika will get a regular bracket sequence () with probability p = frac{3}{4} , and she will get an irregular bracket sequence )( with probability 1 - p = frac{1}{4} . The sought probability is frac{3}{4} , and 249 ,561 ,089 cdot 4 equiv 3 pmod{998 ,244 ,353} . In the second example, the sought probability is frac{11}{25} . '...