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.
from __future__ import division, print_function


def main():
    # Template 1.0
    import sys, re, math
    from collections import deque, defaultdict, Counter, OrderedDict
    from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, floor
    from heapq import heappush, heappop, heapify, nlargest, nsmallest
    def STR():
        return list(input())

    def INT():
        return int(input())

    def MAP():
        return map(int, input().split())

    def LIST():
        return list(map(int, input().split()))

    def list2d(a, b, c):
        return [[c] * b for i in range(a)]

    def sortListWithIndex(listOfTuples, idx):
        return (sorted(listOfTuples, key=lambda x: x[idx]))

    def sortDictWithVal(passedDic):
        temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))
        toret = {}
        for tup in temp:
            toret[tup[0]] = tup[1]
        return toret

    def sortDictWithKey(passedDic):
        return dict(OrderedDict(sorted(passedDic.items())))

    INF = float('inf')
    mod = 10 ** 9 + 7

    t = INT()

    while(t!=0):
        n = INT()

        a = LIST()
        b = LIST()

        if(a[0]==b[0]):
            pos,neg,z = 0,0,0
            if(a[0]>0): pos = 1
            elif(a[0]<0):   neg = 1
            else:   z = 1

            flag = 0

            for i in range(1, n):
                if(a[i]!=b[i]):

                    temp = b[i]-a[i]
                    if(temp>0 and pos==1):
                        if (a[i] > 0):
                            pos = 1
                        elif (a[i] < 0):
                            neg = 1
                        else:
                            z = 1
                        continue
                    elif(temp<0 and neg==1):
                        if (a[i] > 0):
                            pos = 1
                        elif (a[i] < 0):
                            neg = 1
                        else:
                            z = 1
                        continue
                    else:
                        flag = 1
                        break

                else:
                    if(a[i]>0):
                        pos=1
                    elif(a[i]<0):
                        neg = 1
                    else:
                        z = 1

            if(flag==1):
                print("NO")
            else:
                print("YES")
        else:
            print("NO")

        t-=1

######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

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')

# Cout implemented in Python
import sys


class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self


cout = ostream()
endl = '\n'


# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A


if __name__ == "__main__":
    main()