Problem: Uncle Bogdan is in captain Flint's crew for a long time and sometimes gets nostalgic for his homeland. Today he told you how his country introduced a happiness index.

There are n cities and n−1 undirected roads connecting pairs of cities. Citizens of any city can reach any other city traveling by these roads. Cities are numbered from 1 to n and the city 1 is a capital. In other words, the country has a tree structure.

There are m citizens living in the country. A p_i people live in the i-th city but all of them are working in the capital. At evening all citizens return to their home cities using the shortest paths. 

Every person has its own mood: somebody leaves his workplace in good mood but somebody are already in bad mood. Moreover any person can ruin his mood on the way to the hometown. If person is in bad mood he won't improve it.

Happiness detectors are installed in each city to monitor the happiness of each person who visits the city. The detector in the i-th city calculates a happiness index h_i as the number of people in good mood minus the number of people in bad mood. Let's say for the simplicity that mood of a person doesn't change inside the city.

Happiness detector is still in development, so there is a probability of a mistake in judging a person's happiness. One late evening, when all citizens successfully returned home, the government asked uncle Bogdan (the best programmer of the country) to check the correctness of the collected happiness indexes.

Uncle Bogdan successfully solved the problem. Can you do the same?

More formally, You need to check: "Is it possible that, after all people return home, for each city i the happiness index will be equal exactly to h_i".

Input

The first line contains a single integer t (1 ≤ t ≤ 10000) — the number of test cases.

The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^5; 0 ≤ m ≤ 10^9) — the number of cities and citizens.

The second line of each test case contains n integers p_1, p_2, …, p_{n} (0 ≤ p_i ≤ m; p_1 + p_2 + … + p_{n} = m), where p_i is the number of people living in the i-th city.

The third line contains n integers h_1, h_2, …, h_{n} (-10^9 ≤ h_i ≤ 10^9), where h_i is the calculated happiness index of the i-th city.

Next n − 1 lines contain description of the roads, one per line. Each line contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i), where x_i and y_i are cities connected by the i-th road.

It's guaranteed that the sum of n from all test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print YES, if the collected data is correct, or NO — otherwise. You can print characters in YES or NO in any case.

Examples

Input


2
7 4
1 0 1 1 0 1 0
4 0 0 -1 0 -1 0
1 2
1 3
1 4
3 5
3 6
3 7
5 11
1 2 5 2 1
-11 -2 -6 -2 -1
1 2
1 3
1 4
3 5


Output


YES
YES


Input


2
4 4
1 1 1 1
4 1 -3 -1
1 2
1 3
1 4
3 13
3 3 7
13 1 4
1 2
1 3


Output


NO
NO

Note

Let's look at the first test case of the first sample: 

<image>

At first, all citizens are in the capital. Let's describe one of possible scenarios: 

  * a person from city 1: he lives in the capital and is in good mood; 
  * a person from city 4: he visited cities 1 and 4, his mood was ruined between cities 1 and 4; 
  * a person from city 3: he visited cities 1 and 3 in good mood; 
  * a person from city 6: he visited cities 1, 3 and 6, his mood was ruined between cities 1 and 3; 

In total, 
  * h_1 = 4 - 0 = 4, 
  * h_2 = 0, 
  * h_3 = 1 - 1 = 0, 
  * h_4 = 0 - 1 = -1, 
  * h_5 = 0, 
  * h_6 = 0 - 1 = -1, 
  * h_7 = 0. 



The second case of the first test: 

<image>

All people have already started in bad mood in the capital — this is the only possible scenario.

The first case of the second test: 

<image>

The second case of the second test: 

<image>

It can be proven that there is no way to achieve given happiness indexes in both cases of the second test. 

Solution (in PYTHON3):

Solution:
#########			##	##     ##    	  ####  #####  ##     #  ##     #		##
	#			   # #	# #   # #		 #	  #	#	#  # #    #  # #    #	   # #
	#			  #  #	#  ###  #	    #		#	#  #  #   #  #  #   #	  #  #
	#			 #####	#	#	#	   #    ###	#	#  #   #  #  #   #  #    #####
	#			#    #	#		#	   #    # #	#	#  #	# #  #	  # #   #    #  
######### 	   #     # 	#		#		##### #	#####  #	 ##  #	   ##  #     #

"""

PPPPPPP       RRRRRRR		    OOOO	  VV        VV    EEEEEEEEEE
PPPPPPPP      RRRRRRRR         OOOOOO     VV       VV	  EE
PPPPPPPPP     RRRRRRRRR       OOOOOOOO    VV      VV	  EE
PPPPPPPP      RRRRRRRR        OOOOOOOO    VV     VV    	  EEEEEE
PPPPPPP       RRRRRRR         OOOOOOOO    VV    VV        EEEEEEE
PP  		  RRRR			  OOOOOOOO    VV   VV         EEEEEE
PP			  RR  RR          OOOOOOOO    VV  VV          EE
PP			  RR    RR         OOOOOO     VV VV           EE
PP			  RR      RR        OOOO      VVVV            EEEEEEEEEE

"""



"""
 Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys

sys.setrecursionlimit(1000000)
input = sys.stdin.readline
read = lambda: map(int, input().split())
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;

 
def ceil(x):
    if x != int(x):
        x = int(x) + 1
    return x
 
def factorial(x, m):
	val = 1
	while x>0:
		val = (val * x) % m
		x -= 1
	return val

def fact(x):
	val = 1
	while x > 0:
		val *= x
		x -= 1
	return val
    
# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;
 
## gcd function
def gcd(a,b):
    if b == 0:
        return a;
    return gcd(b, a % b);

## lcm function
def lcm(a, b):
	return (a * b) // math.gcd(a, b)
 
## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
	if k > n:
		return 0
	if(k > n - k):
		k = n - k
	res = 1
	for i in range(k):
		res = res * (n - i)
		res = res / (i + 1)
	return int(res)
 
## upper bound function code -- such that e in a[:i] e < x;

 
## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes
 
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 
 
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b;
 
# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];
 
# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;
 
 
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x
 
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    prime[0], prime[1] = False, False
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

# Euler's Toitent Function phi
def phi(n) : 
  
    result = n 
    p = 2
    while(p * p<= n) : 
        if (n % p == 0) : 
            while (n % p == 0) : 
                n = n // p 
            result = result * (1.0 - (1.0 / (float) (p))) 
        p = p + 1
    if (n > 1) : 
        result = result * (1.0 - (1.0 / (float)(n))) 
   
    return (int)(result) 

def is_prime(n):
	if n == 0:
		return False
	if n == 1:
		return True
	for i in range(2, int(n ** (1 / 2)) + 1):
		if not n % i:
			return False
 
	return True
 
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array
 
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
    res = []
    for i in range(2, int(x ** 0.5) + 1):
    	while x % i == 0:
    		res.append(i)
    		x //= i
    if x != 1:
   		res.append(x)
    return res
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

def factors(n):
	res = []
	for i in range(1, int(n ** 0.5) + 1):
		if n % i == 0:
			res.append(i)
			res.append(n // i)
	return list(set(res))
 
## taking integer array input
def int_array():
    return list(map(int, input().strip().split()));
 
def float_array():
    return list(map(float, input().strip().split()));
 
## taking string array input
def str_array():
    return input().strip().split();

def binary_search(low, high, w, h, n):
	while low < high:
		mid = low + (high - low) // 2
		# print(low, mid, high)
		if check(mid, w, h, n):
			high = mid
		else:
			low = mid + 1
	return low

## for checking any conditions
def check(time, x, y, n):
	time -= min(x, y)
	if time < 0:
		return False
	# print(copies_in_l_seconds, l, time, c)
	return time // x + time // y + 1 >= n
 
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
 
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
 
from itertools import permutations
import math
import bisect as bis
import random
import sys

def dfs(x, par):
	global f
	tot[x] = p[x]
	sumgd = 0
	for v in g[x]:
		if v != par:
			dfs(v, x)
			tot[x] += tot[v]
			sumgd += gd[v]
	if (tot[x] + h[x]) % 2 == 1:
		f = 1
	gd[x] = (tot[x] + h[x]) // 2
	if gd[x] < 0 or gd[x] > tot[x]: 
		f = 1
	if sumgd > gd[x]:
		f = 1

	
for _ in range(int(input())):
	n,m=map(int,input().split())
	p=[-1]+list(map(int,input().split()))
	h=[-1]+list(map(int,input().split()))
	g=[[] for i in range(n+1)]
	f=0
	for i in range(n-1):	
		x,y=map(int,input().split())
		g[x].append(y)
		g[y].append(x)
	tot=[0]*(n+1)
	gd=[0]*(n+1)
	dfs(1,0)
	if f:
		print('NO')
	else:
		print("YES")

	# fin_time = datetime.now()
# 	print("Execution time (for loop): ", (fin_time-init_time))