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 $t$ ($1 \le t \le 10000$) — the number of test cases.

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

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

The third line contains $n$ integers $h_1, h_2, \ldots, h_{n}$ ($-10^9 \le h_i \le 10^9$), 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 $y_i$ ($1 \le x_i, y_i \le n$; $x_i \neq y_i$), 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 $2 \cdot 10^5$.


-----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:  $\text{of}_{0}$ 

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

The first case of the second test:  $\text{of} 0$ 

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.

Difficulty: interview

Starter code:

Solutions:
import sys
input = sys.stdin.readline
T = int(input())



for testcase in range(1,T+1):
    n,m = map(int,input().split())
    p = tuple(map(int,input().split()))
    h = tuple(map(int,input().split()))

    a = [0]*n
    b = [0]*n

    edge = [[] for i in range(n)]
    for _ in range(n-1):
        x,y = map(int,input().split())
        edge[x-1].append(y-1)
        edge[y-1].append(x-1)
    
    par = [-1]*n
    tank = [0]
    order = []
    while tank:
        now = tank.pop()
        order.append(now)
        for e in edge[now]:
            if par[now] != e:
                par[e] = now
                tank.append(e)
    flag = True
    for e in order[::-1]:
        if (a[e]-b[e]-p[e] <= h[e] <= a[e]+b[e]+p[e]) and (h[e]+a[e]+b[e]+p[e])%2 == 0:
            if e != 0:
                a[par[e]] += (h[e]+a[e]+b[e]+p[e])//2
                b[par[e]] += (h[e]+a[e]+b[e]+p[e])//2 - h[e]
        else:
            flag = False
            break
    if flag:
        print("YES")
    else:
        print("NO")


from sys import stdin
import sys
sys.setrecursionlimit(300000)

def dfs(v,pa):

    good = 0
    bad = 0
    for nex in lis[v]:
        if nex != pa:
            nans,ng,nb = dfs(nex,v)
            if not nans:
                return nans,0,0
            good += ng
            bad  += nb

    num = good + bad + p[v]
    if (num - h[v]) % 2 == 0:
        newbad  = (num - h[v])//2
    else:
        return False,0,0
    newgood = num - newbad

    if newbad - p[v] > bad or newgood < good or newbad < 0 or newgood < 0:
        return False,0,0
    else:
        return True,newgood,newbad




tt = int(stdin.readline())

for loop in range(tt):

    n,m = list(map(int,stdin.readline().split()))
    p = list(map(int,stdin.readline().split()))
    h = list(map(int,stdin.readline().split()))

    lis = [ [] for i in range(n)]
    for i in range(n-1):
        v,u = list(map(int,stdin.readline().split()))
        v -= 1
        u -= 1
        lis[v].append(u)
        lis[u].append(v)

    ans,good,bad = dfs(0,0)
    
    if ans:
        print ("YES")
    else:
        print ("NO")


import sys
input = sys.stdin.readline


def topological_sort(tree: list, root) -> list:
    n = len(tree)
    visited = [False] * n
    visited[root] = True
    tp_sorted = [root]
    stack = [root]
    while stack:
        v = stack.pop()
        for nxt_v in tree[v]:
            if visited[nxt_v]:
                continue
            visited[nxt_v] = True
            stack.append(nxt_v)
            tp_sorted.append(nxt_v)
    return tp_sorted


def get_par(tree: list, root: int) -> list:
    n = len(tree)
    visited = [False] * n
    visited[root] = True
    par = [-1] * n
    stack = [root]
    while stack:
        v = stack.pop()
        for nxt_v in tree[v]:
            if visited[nxt_v]:
                continue
            visited[nxt_v] = True
            stack.append(nxt_v)
            par[nxt_v] = v
    return par


t = int(input())
for _ in range(t):
    n, m = map(int, input().split())
    p = list(map(int, input().split()))
    h = list(map(int, input().split()))
    edges = [list(map(int, input().split())) for i in range(n - 1)]
    
    tree = [[] for i in range(n)]
    for a, b in edges:
        a -= 1
        b -= 1
        tree[a].append(b)
        tree[b].append(a)

    root = 0
    tp_sort = topological_sort(tree, root)
    par = get_par(tree, root)
    cnts = [[0, 0] for i in range(n)]
    flag = True
    for v in tp_sort[::-1]:
        good, bad = 0, 0
        for nxt_v in tree[v]:
            if nxt_v == par[v]:
                continue
            else:
                good += cnts[nxt_v][0]
                bad += cnts[nxt_v][1]

        sum_p = good + bad + p[v]
        if (sum_p + h[v]) % 2 != 0:
            flag = False
            break
        g = (sum_p + h[v]) // 2
        b = (sum_p - h[v]) // 2
        
        inc_g = g - good
        if 0 <= inc_g <= p[v] + bad:
            cnts[v] = g, b
            continue
        else:
            flag = False
            break

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

import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write('\n'.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
from fractions import Fraction
#sys.setrecursionlimit(100000)
INF = float('inf')
mod = int(1e9)+7

def dfs(graph, start=0):
    n = len(graph)

    dp = [0] * n
    visited, finished = [False] * n, [False] * n
    par = [0]*n

    stack = [start]
    while stack:
        start = stack[-1]

        # push unvisited children into stack
        if not visited[start]:
            visited[start] = True
            for child in graph[start]:
                if not visited[child]:
                    stack.append(child)
                    par[child]=start

        else:
            stack.pop()

            # base case
            dp[start] += p[start]

            # update with finished children
            for child in graph[start]:
                if finished[child]:
                    dp[start] += dp[child]

            finished[start] = True

    return dp

def dfs2(graph, start=0):
    n = len(graph)

    dp = [0] * n
    visited, finished = [False] * n, [False] * n
    n = [0]*n

    stack = [start]
    while stack:
        start = stack[-1]

        # push unvisited children into stack
        if not visited[start]:
            visited[start] = True
            for child in graph[start]:
                if not visited[child]:
                    stack.append(child)

        else:
            stack.pop()

            # base case
            dp[start] += (v[start]+h[start])//2
            a=v[start]-dp[start]
            if dp[start]-a!=h[start]  or a<0:
                return False

            # update with finished children
            k=0
            for child in graph[start]:
                if finished[child]:
                    k += dp[child]
            if k>dp[start]:
                return False

            finished[start] = True

    return True

for t in range(int(data())):
    n,m=mdata()
    p=mdata()
    h=mdata()
    g=[[] for i in range(n)]
    for i in range(n-1):
        u,v=mdata()
        u,v=u-1,v-1
        g[u].append(v)
        g[v].append(u)
    v=dfs(g,0)
    if dfs2(g,0):
        out("YES")
    else:
        out("NO")