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Easy Problem Once More

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 109    Accepted Submission(s): 4


Problem Description
Define matrix , if for every m(1 ≒ m ≒ n),

then matrix A can be called as partially negative matrix. Here matrix
, and {i1,..,im} is a sub set of {1,..,n}.If you are not familiar with determinant of a matrix, please read the Note part of this problem.
For example, matrix is a partially negative matrix because |-2|, |-6| andare negative.
A symmetric matrix is a square matrix that equals to its transpose. Formally, matrix A is symmetric if A = AT. For example, is a symmetric matrix.
Given two N-dimensional vector x and b, and we guarantee that there will be at least
one 0 value in vector b. You task is to judge if there exists a symmetric partially
negative matrix A, which fulfills Ax = b.
 

Input
There are several test cases. Proceed to the end of file.
Each test case is described in three lines.
The first line contains one integer N (2 ≒ N ≒ 100000) .
The second line contains N integers xi (-1000000 < xi < 1000000, 1 ≒ i ≒ N), which is vector x.
The third line contains N integers bi (-1000000 < bi < 1000000, 1 ≒ i ≒ N), which is vector b. There will be at least one bi which equals to zero.
 

Output
For each test case, output ※Yes§ if there exists such a matrix A, or ※No§ if there is no such matrix.
 

Sample Input
2 2 1 0 6
 

Sample Output
Yes
 

Hint
There exists a symmetric partially negative matrix


Note

Determinant of an n ℅ n matrix A is defined as below:

Here the sum is computed over all permutations 考 of the set {1, 2, ..., n}.
A permutation is a function that reorders this set of integers.
The value in the ith position after the reordering 考 is denoted 考i.
For example, for n = 3, the original sequence 1, 2, 3 might be reordered to 考 = [2, 3, 1],with 考1 = 2, 考2 = 3, and 考3 = 1.
The set of all such permutations (also known as the symmetric group on n elements) is denoted Sn.
For each permutation 考, sgn(考) denotes the signature of 考, a value that is +1
whenever the reordering given by 考 can be achieved by successively interchanging two entries an even number of times,
and &#8722;1 whenever it can be achieved by an odd number of such interchanges.
 

Source
 

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