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 bisect import *
from collections import *
from itertools import *
import functools
import sys
import math
from decimal import *
from copy import *
from heapq import *
from fractions import *
getcontext().prec = 30
MAX = sys.maxsize
MAXN = 300010
MOD = 10**9+7
spf = [i for i in range(MAXN)]
spf[0]=spf[1] = -1
def sieve():
    for i in range(2,MAXN,2):
        spf[i] = 2
    for i in range(3,int(MAXN**0.5)+1):
        if spf[i]==i:
            for j in range(i*i,MAXN,i):
                if spf[j]==j:
                    spf[j]=i
def fib(n,m):
    if n == 0:
        return [0, 1]
    else:
        a, b = fib(n // 2)
        c = ((a%m) * ((b%m) * 2 - (a%m)))%m
        d = ((a%m) * (a%m))%m + ((b)%m * (b)%m)%m
        if n % 2 == 0:
            return [c, d]
        else:
            return [d, c + d]

def charIN(x= ' '):
    return(sys.stdin.readline().strip().split(x))

def arrIN(x = ' '):
    return list(map(int,sys.stdin.readline().strip().split(x)))

def ncr(n,r):
    num=den=1
    for i in range(r):
        num = (num*(n-i))%MOD
        den = (den*(i+1))%MOD

    return (num*(pow(den,MOD-2,MOD)))%MOD

def flush():
    return sys.stdout.flush()

'''*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*'''
def solve():
    n = int(input())
    a = arrIN()
    b = arrIN()
    x = [[0,0,0] for i in range(n)]
    for i in range(n):
        x[i][0] = int(a[i]==-1)
        x[i][1] = int(a[i]==0)
        x[i][2] = int(a[i]==1)
        x[i][0]|=x[i-1][0]
        x[i][1]|=x[i-1][1]
        x[i][2]|=x[i-1][2]
    if a[0]!=b[0]:
        print('NO')
    else:
        for i in range(1,n):
            if a[i]!=b[i]:
                if a[i]>b[i]:
                    if not x[i-1][0]:
                        print('NO')
                        break
                else:
                    if not x[i-1][2]:
                        print('NO')
                        break
        else:
            print('YES')


t = int(input())
for i in range(t):
    solve()

