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>).
line = input()
pieces=line.split()
n= int(pieces[0])
q = int(pieces[1])#num of queries

line2= input()
pieces2= line2.split()
colorList=[]
for i in range(len(pieces2)):
    colorList.append({i+1:pieces2[i]})

edgeList=[]
for i in range(n):
    edgeList.append([])
for i in range(n-1):
    line = input()
    edge = line.split()
    edge1 = int(edge[0])
    edge2 = int(edge[1])
    edgeList[(edge1-1)].append(edge2)
    edgeList[(edge2-1)].append(edge1)


parentList = []

for i in range(n):
    parentList.append(0)


def DFS(p,parent):
    for v in edgeList[p-1]:
        if v!=parent:
            DFS(v,p)
            parentList[v-1]=p

def DFS2(p,parent):
    for v in edgeList[p-1]:
        if v!=parent:
            colorCnt[int(colorList[v-1][v])-1] +=1
            DFS2(v,p)

DFS(1,0)


global kj
for i in range(q):
    lineQ= input()
    query = lineQ.split()
    vj = int(query[0]) #start vertax
    kj = int(query[1]) #least num same color
    colorCnt=[]
    for i in range(n):
        colorCnt.append(0)
    colorCnt[int(colorList[vj-1][vj])-1] +=1
    DFS2(vj,parentList[vj-1])
    count = 0
    for i in colorCnt:
        if i >= kj:
            count +=1
    print (count)

