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:
n = int(input())

from collections import defaultdict
import math


def compute_helper(TOT_POPS, graph, pop_cities):
    children = defaultdict(list)
    STACK = []
    VISITED = {}
    VISITED[1] = True
    initial_city = 1
    STACK.append(initial_city)
    while (len(STACK)) != 0:
        curr_city = STACK[-1]
        if (len(graph[curr_city]) == 1) and (curr_city != 1):
            TOT_POPS[curr_city] = pop_cities[curr_city]
            STACK.pop()
            continue

        for child in graph[curr_city]:
            if child not in VISITED:
                STACK.append(child)
                children[curr_city].append(child)
                VISITED[child] = True

        if all([child in TOT_POPS for child in children[curr_city]]):
            STACK.pop()
            TOT_POPS[curr_city] = pop_cities[curr_city] + sum([TOT_POPS[child] for child in children[curr_city]])


def compute_total_population(marker, TOT_POPS, city, graph, pop_cities):
    total_population = pop_cities[city]
    marker[city] = True
    for child_city in graph[city]:
        if child_city in marker:
            continue
        if len(graph[child_city]) == 1:  # in other words this is leaf
            total_population += pop_cities[child_city]
            TOT_POPS[child_city] = pop_cities[child_city]
        else:
            total_population += compute_total_population(marker, TOT_POPS, child_city, graph, pop_cities)
    TOT_POPS[city] = total_population
    return total_population


def non_recursive_explore(TOT_POPS, happiness_by_city, graph):
    STACK = [1]
    VISITED = {}
    VISITED[1] = True
    children = defaultdict(list)
    NB_HAPPY = {}
    while len(STACK) != 0:
        curr_city = STACK[-1]

        if (TOT_POPS[curr_city] - happiness_by_city[curr_city]) % 2 == 0 and abs(happiness_by_city[curr_city]) <= \
                TOT_POPS[curr_city]:
            pass
        else:
            return False

        # if city is a leaf, there is nothing to verify
        if len(graph[curr_city]) == 1 and curr_city != 1:
            NB_HAPPY[curr_city] = int((TOT_POPS[curr_city] + happiness_by_city[curr_city]) / 2)
            STACK.pop()
            continue
        # If city is not a leaf, we try to add its children to the stack
        for child_city in graph[curr_city]:
            if child_city not in VISITED:
                VISITED[child_city] = True
                children[curr_city].append(child_city)
                STACK.append(child_city)
        # If the current city's children have been visited, then we can make a judgement on the current city
        if all([child in NB_HAPPY for child in children[curr_city]]):
            nb_happy_person_in_children = sum([NB_HAPPY[child] for child in children[curr_city]])
            nb_happy_person_in_parent = (happiness_by_city[curr_city] + TOT_POPS[curr_city]) // 2
            nb_unhappy_people = (TOT_POPS[curr_city] - happiness_by_city[curr_city]) // 2
            nb_of_happy_person_left_behind = population_by_city[curr_city] - nb_unhappy_people
            nb_of_happy_person_left_behind = nb_of_happy_person_left_behind if nb_of_happy_person_left_behind > 0 else 0

            # The main condition is here : At each level we need to verify that the number of happy person is increasing with regards to the parent
            if nb_happy_person_in_children <= nb_happy_person_in_parent - nb_of_happy_person_left_behind:
                NB_HAPPY[curr_city] = nb_happy_person_in_parent
                STACK.pop()
                continue
            else:
                return False
    return True

__import__('sys').setrecursionlimit(5056)
def explore(city, marker, TOT_POPS, happiness_by_city, graph):
    marker[city] = True
    nb_happy_person_in_parent = int((TOT_POPS[city] + happiness_by_city[city]) / 2)
    nb_happy_person_in_children = 0
    # if city is a leaf, there is nothing to verify
    if len(graph[city]) == 1 and city != 1:
        if (TOT_POPS[city] - happiness_by_city[city]) % 2 == 0:
            return True
        else:
            return False

    for child_city in graph[city]:
        try:
            if child_city not in marker:
                if abs(happiness_by_city[child_city]) > TOT_POPS[child_city]:
                    return False
                nb_happy_person_in_children += int((TOT_POPS[child_city] + happiness_by_city[child_city]) / 2)
                VALID = explore(child_city, marker, TOT_POPS, happiness_by_city, graph)
                if not VALID:
                    return False
        except Exception as e:
            print(traceback.format_exc())

    # The main condition is here : At each level we need to verify that the number of happy person is increasing with regards to the parent
    nb_of_happy_person_left_behind = population_by_city[city] - int(
        (TOT_POPS[city] - happiness_by_city[city]) / 2) if \
        population_by_city[city] - int((TOT_POPS[city] - happiness_by_city[city]) / 2) > 0 else 0
    if (nb_happy_person_in_children <= nb_happy_person_in_parent - nb_of_happy_person_left_behind) and (abs(
            happiness_by_city[city]) <= TOT_POPS[city]) and (TOT_POPS[city] - happiness_by_city[city]) % 2 == 0:
        return True
    else:
        return False


for _ in range(n):
    TOT_POPS = {}
    graph = defaultdict(list)
    nb_cities, nb_citizens = map(int, input().split())
    population_by_city = [0] + list(map(int, input().split()))
    happiness_by_city = [0] + list(map(int, input().split()))
    for _ in range(nb_cities - 1):
        city1, city2 = map(int, input().split())
        graph[city1].append(city2)
        graph[city2].append(city1)
    # We compute the total population for each city one time through DFS
    marker0 = {}
    import traceback

    try:
        compute_helper(TOT_POPS, graph, population_by_city)
        # compute_total_population(marker0, TOT_POPS, 1, graph, population_by_city)
    except Exception as e:
        print(e)
        print(traceback.format_exc())
        exit(1)


    try:
        marker = {}
        if explore(1, marker, TOT_POPS, happiness_by_city, graph):
            print('YES')
        else:
            print('NO')
    except Exception as e:
        print(e)

    """
    try:
        if non_recursive_explore(TOT_POPS, happiness_by_city, graph):
            print('YES')
        else:
            print('NO')
    except Exception as e:
        print(e)
    """