Problem:
Score : 600 points

Problem Statement

Given are a prime number p and a sequence of p integers a_0, \ldots, a_{p-1} consisting of zeros and ones.

Find a polynomial of degree at most p-1, f(x) = b_{p-1} x^{p-1} + b_{p-2} x^{p-2} + \ldots + b_0, satisfying the following conditions:

For each i (0 \leq i \leq p-1), b_i is an integer such that 0 \leq b_i \leq p-1.

For each i (0 \leq i \leq p-1), f(i) \equiv a_i \pmod p.

Constraints

2 \leq p \leq 2999

p is a prime number.

0 \leq a_i \leq 1

Input

Input is given from Standard Input in the following format:

p
a_0 a_1 \ldots a_{p-1}

Output

Print b_0, b_1, \ldots, b_{p-1} of a polynomial f(x) satisfying the conditions, in this order, with spaces in between.

It can be proved that a solution always exists. If multiple solutions exist, any of them will be accepted.

Sample Input 1

2
1 0

Sample Output 1

1 1

f(x) = x + 1 satisfies the conditions, as follows:

f(0) = 0 + 1 = 1 \equiv 1 \pmod 2

f(1) = 1 + 1 = 2 \equiv 0 \pmod 2

Sample Input 2

3
0 0 0

Sample Output 2

0 0 0

f(x) = 0 is also valid.

Sample Input 3

5
0 1 0 1 0

Sample Output 3

0 2 0 1 3

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

require 'pp'
require 'matrix'

p = gets.chomp!.to_i
a_n = gets.chomp!.split(" ").map(&:to_i)

mat = []
mat[0] = [0] * p
mat[0][0] = 1
mat[1] = [1] * p
2.upto(p-1) do |i|
    mat << 0.upto(p-1).map {|j| (i ** j) % p }
end
matrix = Matrix[*mat]
a_n_mat = Matrix.column_vector(a_n)

pp matrix
pp matrix.inverse
pp a_n_mat

b_n = matrix * a_n_mat
puts b_n.t.to_a.flatten.map {|i| i % p }.join(" ")