You have a rooted tree consisting of *n* vertices. Each vertex of the tree has some color. We will assume that the tree vertices are numbered by integers from 1 to *n*. Then we represent the color of vertex *v* as *c*<sub class="lower-index">*v*</sub>. The tree root is a vertex with number 1.

In this problem you need to answer to *m* queries. Each query is described by two integers *v*<sub class="lower-index">*j*</sub>, *k*<sub class="lower-index">*j*</sub>. The answer to query *v*<sub class="lower-index">*j*</sub>, *k*<sub class="lower-index">*j*</sub> is the number of such colors of vertices *x*, that the subtree of vertex *v*<sub class="lower-index">*j*</sub> contains at least *k*<sub class="lower-index">*j*</sub> vertices of color *x*.

You can find the definition of a rooted tree by the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory).
The first line contains two integers *n* and *m* (2 ≤ *n* ≤ 10<sup class="upper-index">5</sup>; 1 ≤ *m* ≤ 10<sup class="upper-index">5</sup>). The next line contains a sequence of integers *c*<sub class="lower-index">1</sub>, *c*<sub class="lower-index">2</sub>, ..., *c*<sub class="lower-index">*n*</sub> (1 ≤ *c*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>). The next *n* - 1 lines contain the edges of the tree. The *i*-th line contains the numbers *a*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*i*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*i*</sub> ≤ *n*; *a*<sub class="lower-index">*i*</sub> ≠ *b*<sub class="lower-index">*i*</sub>) — the vertices connected by an edge of the tree.

Next *m* lines contain the queries. The *j*-th line contains two integers *v*<sub class="lower-index">*j*</sub>, *k*<sub class="lower-index">*j*</sub> (1 ≤ *v*<sub class="lower-index">*j*</sub> ≤ *n*; 1 ≤ *k*<sub class="lower-index">*j*</sub> ≤ 10<sup class="upper-index">5</sup>).
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
# from math import *
# from bisect import *
# from heapq import *
# from collections import defaultdict as dd
# from collections import OrderedDict as odict
# from collections import Counter as cc
# from collections import deque
sys.setrecursionlimit(10**5+100) #this is must for dfs
# ______________________________________________________________________________________________________
# segment tree for range minimum query
# n = int(input())	
# a = list(map(int,input().split()))
# st = [float('inf') for i in range(4*len(a))]
# def build(a,ind,start,end):
# 	if start == end:
# 		st[ind] = a[start]
# 	else:
# 		mid = (start+end)//2
# 		build(a,2*ind+1,start,mid)
# 		build(a,2*ind+2,mid+1,end)
# 		st[ind] = min(st[2*ind+1],st[2*ind+2])
# build(a,0,0,n-1)
# def query(ind,l,r,start,end):
# 	if start>r or end<l:
# 		return float('inf')
# 	if l<=start<=end<=r:
# 		return st[ind]
# 	mid = (start+end)//2
# 	return min(query(2*ind+1,l,r,start,mid),query(2*ind+2,l,r,mid+1,end))
# ______________________________________________________________________________________________________
# Checking prime in O(root(N))
# def isprime(n):
#     if (n % 2 == 0 and n > 2) or n == 1: return 0
#     else:
#         s = int(n**(0.5)) + 1
#         for i in range(3, s, 2):
#             if n % i == 0:
#                 return 0
#         return 1
# def lcm(a,b):
#   return (a*b)//gcd(a,b)
# ______________________________________________________________________________________________________
# nCr under mod
# def C(n,r,mod):
#   if r>n:
#       return 0
#   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
# ______________________________________________________________________________________________________
# For smallest prime factor of a number
# M = 2*(10**5)+10
# pfc = [i for i in range(M)]
# def pfcs(M):
#   for i in range(2,M):
#       if pfc[i]==i:
#           for j in range(i+i,M,i):
#               if pfc[j]==j:
#                   pfc[j] = i
#   return
# pfcs(M)
# ______________________________________________________________________________________________________
tc = 1
# tc, = inp()
for _ in range(tc):
	n,m = inp()
	c = [0]+inp()
	g = [[] for i in range(n+1)]
	BLK = int((2*n)**0.5)+1
	class Node:
		def __init__(self,start,end,k,idx):
			self.s = start
			self.t = end
			self.k = k
			self.idx = idx
	for i in range(n-1):
		a,b = inp()
		g[a].append(b)
		g[b].append(a)
	ft = [-1]*(2*n+10)
	s = [-1]*(n+1)
	t = [-1]*(n+1)
	timer = 1
	def dfs(node,par):
		global timer
		ft[timer] = node
		s[node] = timer
		timer+=1
		for to in g[node]:
			if to!=par:
				dfs(to,node)
		ft[timer] = node
		t[node] = timer
		timer+=1
	dfs(1,-1)
	q = []
	for i in range(m):
		a,b = inp()
		q.append(Node(s[a],t[a],b,i))
	q.sort(key = lambda x:(x.s//BLK,x.t))
	freq = [0]*(n+1)
	fr = [0]*(10**5+1)
	freq[0] = n
	nodes = [0]*(n+1)
	ans = [0]*m
	def add(node):
		node = ft[node]
		nodes[node]+=1
		if nodes[node]==2:
			fr[c[node]]+=1
			freq[fr[c[node]]]+=1
	def remove(node):
		node = ft[node]
		nodes[node]-=1
		if nodes[node]==1:
			freq[fr[c[node]]]-=1
			fr[c[node]]-=1
	l,r = 1,0
	for i in q:
		while(l<i.s):
			remove(l)
			l+=1
		while(l>i.s):
			l-=1
			add(l)
		while(r<i.t):
			r+=1
			add(r)
		while(i.t<r):
			remove(r)
			r-=1
		ans[i.idx] = freq[i.k]
	print(*ans,sep = '\n')