Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:

There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.

Anton can perform the following sequence of operations any number of times:

  1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once. 
  2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j. 



For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.

Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?

Input

Each test contains multiple test cases. 

The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.

The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.

The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.

It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.

Output

For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.

You can print each letter in any case (upper or lower).

Example

Input


5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1


Output


YES
NO
YES
YES
NO

Note

In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]

In the second test case we can't make equal numbers on the second position.

In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.

In the last lest case, it is impossible to make array a equal to the array b.
"""Template for Python Competitive Programmers prepared by Mayank Chaudhary """

# to use the print and division function of Python3
from __future__ import division, print_function

"""value of mod"""
MOD = 998244353
mod = 10**9 + 7

"""use resource"""
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

"""for factorial"""

# def prepare_factorial():
#     fact = [1]
#     for i in range(1, 1000005):
#         fact.append((fact[-1] * i) % mod)
#     ifact = [0] * 1000005
#     ifact[1000004] = pow(fact[1000004], mod - 2, mod)
#     for i in range(1000004, 0, -1):
#         ifact[i - 1] = (i * ifact[i]) % mod
#
#     return fact, ifact

"""uncomment next 4 lines while doing recursion based question"""
import threading
threading.stack_size(1<<27)
import sys
sys.setrecursionlimit(10000)


"""uncomment modules according to your need"""
# from bisect import bisect_left, bisect_right, insort
# import itertools
# from math import floor, ceil, sqrt, degrees, atan, pi
# from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
# from collections import Counter, defaultdict, deque
# from copy import deepcopy
'''
def modinv(n, p):
    return pow(n, p - 2, p)
'''


# def ncr(n, r,  fact, ifact):  # for using this uncomment the lines calculating fact and ifact
#     t = (fact[n] * (ifact[r]*ifact[n-r]) % mod)% mod
#     return t



def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()


# def GCD(x, y):
#     while (y):
#         x, y = y, x % y
#     return x
#
# def lcm(x, y):
#     return (x*y)//(GCD(x, y))

# def get_xor(n):
#     return [n,1,n+1,0][n%4]

# def binary_expo(a, b):
#
#     result = 1
#     while b:
#         if b&1:
#             result *= a
#             b-=1
#         else:
#             a *= a
#             b >>= 1
#     return result



"""*******************************************************"""
def all_factors(n):
    """returns a sorted list of all distinct factors of n"""
    small, large = [], []
    for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
        if not n % i:
            small.append(i)
            large.append(n // i)
    if small[-1] == large[-1]:
        large.pop()
    large.reverse()
    small.extend(large)
    return small

def main():

    T = int(input())
    while T:
        n = int(input())
        A = get_array()
        B = get_array()

        mydict = dict()
        for i in A:
            mydict[i] = mydict.get(i, 0)+1

        flag = 0
        for i in range(n-1, -1, -1):

            if mydict[A[i]]>1:
                mydict[A[i]] -= 1
            else:
                del mydict[A[i]]

            need = B[i] - A[i]
            if need==0:
                continue
            elif need>0:
                if 1 not in mydict:
                    flag = 1
                    break
            else:
                if -1 not in mydict:
                    flag = 1
                    break
        if flag:
            print('NO')
        else:
            print('YES')
        T-=1




""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

""" main function"""

if __name__ == '__main__':
    # main()
    threading.Thread(target=main).start()