Circulant Matrix
题号:NC17637
时间限制:C/C++/Rust/Pascal 1秒,其他语言2秒
空间限制:C/C++/Rust/Pascal 256 M,其他语言512 M
64bit IO Format: %lld

题目描述

Niuniu has recently learned how to use Gaussian elimination to solve systems of linear equations.
Given n and a[i], where n is a power of 2, let's consider an n x n matrix A.

The index of A[i][j] and a[i] are numbered from 0.
The element A[i][j] satisfies A[i][j] = a[i xor j],
https://en.wikipedia.org/wiki/Bitwise_operation#XOR

Let p = 1000000007.
Consider the equation 
A x = b (mod p)
where A is an n x n matrix, and x and b are both n x 1 row vector.

Given n, a[i], b[i], you need to solve the x.
For example, when n = 4, the equations look like
A[0][0]*x[0] + A[0][1]*x[1] + A[0][2]*x[2] + A[0][3]*x[3] = b[0] (mod p)
A[1][0]*x[0] + A[1][1]*x[1] + A[1][2]*x[2] + A[1][3]*x[3] = b[1] (mod p)
A[2][0]*x[0] + A[2][1]*x[1] + A[2][2]*x[2] + A[2][3]*x[3] = b[2] (mod p)
A[3][0]*x[0] + A[3][1]*x[1] + A[3][2]*x[2] + A[3][3]*x[3] = b[3] (mod p)
and the matrix A can be decided by the array a.

It is guaranteed that there is a unique solution x for these equations.

输入描述:

The first line contains an integer, which is n.
The second line contains n integers, which are the array a.
The third line contains n integers, which are the array b.

1 <= n <= 262144
p = 1000000007
0 <= a[i] < p
0 <= b[i] < p

输出描述:

The output should contains n lines.
The i-th(index from 0) line should contain x[i].
x[i] is an integer, and should satisfy 0 <= x[i] < p.
示例1

输入

复制
4
1 10 100 1000
1234 2143 3412 4321

输出

复制
4
3
2
1