C. Kefa and Park

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Kefa decided to celebrate his first big salary by going to the restaurant.

He lives by an unusual park. The park is a rooted tree consisting of n vertices with the root at vertex 1.
Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices
with cats in them.

The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than m consecutive vertices
with cats.

Your task is to help Kefa count the number of restaurants where he can go.

Input

The first line contains two integers, n and m (2 ≤ n ≤ 105, 1 ≤ m ≤ n)
— the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.

The second line contains n integers a1, a2, ..., an,
where each ai either
equals to 0 (then vertex i has
no cat), or equals to 1 (then vertex i has
a cat).

Next n - 1 lines contains the edges of the tree in the format "xi yi"
(without the quotes) (1 ≤ xi, yi ≤ n, xi ≠ yi),
where xiand yi are
the vertices of the tree, connected by an edge.

It is guaranteed that the given set of edges specifies a tree.

Output

A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most mconsecutive vertices
with cats.

Sample test(s)

input

4 1
1 1 0 0
1 2
1 3
1 4

output

2

input

7 1
1 0 1 1 0 0 0
1 21 3
2 4
2 5
3 6
3 7

output

2

Note

Let us remind you that a tree is a connected graph on n vertices and n - 1 edge.
A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex
is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children.

Note to the first sample test:

The vertices containing cats are marked red. The restaurants
are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2.

Note to the second sample test:

The restaurants are located at vertices 4, 5, 6, 7. Kefa
can't go to restaurants 6, 7.

解题说明：此题是图论题，建一棵树，然后对树进行深度优先搜索题，由于一条路上不能连续出现k只猫，遍历路径时进行判断。如果走到了叶子那就得到了满足条件的一条路径。

#include<cstdio>
#include <cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

int n,m,x,y,cnt,a[100005];
vector<int> b[100005];

void dfs(int p,int q,int pr)
{
int i,k,sz = b[p].size();
if(q > m)
{
return;
}
if(sz == 1 && p != 1)
{
cnt++;
return;
}
for(i=0; i<sz; i++)
{
k = b[p][i];
if(k == pr)
{
continue;
}
if(a[k])
{
dfs(k,q+1,p);
}
else
{
dfs(k,0,p);
}
}
}

int main()
{
int i;
cin>>n>>m;
for(i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}
for(i=1; i<n; i++)
{
scanf("%d%d",&x,&y);
b[x].push_back(y);
b[y].push_back(x);
}
dfs(1,a[1],0);
cout<<cnt<<endl;
return 0;
}