Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i.  Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0.  Now, the following Q operations will be performed:    * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j.  Find the value of the counter on each vertex after all operations.
def resolve():
	from collections import deque
	
	n, q = map(int, input().split())
	adj = [[] for i in range(n+1)]
	for i in range(n-1):
		a, b = map(int, input().split())
		adj[a].append(b)
		adj[b].append(a)
	p = [list(map(int, input().split())) for i in range(q)]
	
	ans = [0 for i in range(n+1)]
	tmp = [0 for i in range(n+1)]
	for i in range(q):
		tmp[p[i][0]] += p[i][1]

	for start, point in enumerate(tmp):
		q = deque([start])
		while q:
			node = q.popleft()
			ans[node] += tmp[start]
			for i in list(filter(lambda x: x>node, adj[node])):
				q.append(i)

	for i in ans[1:]:
		print(i, end=" ")

resolve()
import sys
import io

sys.setrecursionlimit(10 ** 6)
def input():
    return sys.stdin.readline()[:-1]
    
def resolve():
	from collections import deque
	
	n, q = map(int, input().split())
	adj = [[] for i in range(n+1)]
	for i in range(n-1):
		a, b = map(int, input().split())
		adj[a-1].append(b-1)
		adj[b-1].append(a-1)
	ans = [0 for i in range(n)]
	for i in range(q):
		p, x = map(int, input().split())
		p -= 1
		ans[p] += x
#	print(f"{adj=}\n{ans=}")

	def dfs(now, prv):
		for next in adj[now]:
#			print(f"{now=}, {next=}")
			if next == prv:
				continue
			ans[next] += ans[now]
			dfs(next, now)

	dfs(0, 0)
	print(*ans)

resolve()
