codeforces 580C Kefa and Park （树上DFS）

Description：

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 xi and 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 m consecutive vertices with cats.

Examples

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 2 1 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.

主人公要到公园去玩，这个公园是一个有根树，主人公起点在点1处，其叶子节点是餐厅所在地，其中红色标记的点都是a【i】==1的点，同时表示这个点有猫.主人公害怕猫，不希望经过连续m个点都有猫的路径，问主人公可以到达哪几个餐厅。

加个标记DFS就行。

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#include<math.h>
using namespace std;
typedef long long ll;
typedef double ld;
const int INF = 0x3f3f3f3f;
const int N=1000010;
int i,j,k;
int m,n,x;
int p,q;
int res,ans,cnt,temp;
int a[N];
vector <int> G[N];
bool vis[N];

void dfs(int x,int cnt)
{
    if (cnt>m)
    {
        return;
    }
    vis[x]=1;
    bool flag=0;     //若没有下一个节点，则本节点是叶节点
    for (int i=0; i<G[x].size(); i++)
    {
        int v=G[x][i];
        if (!vis[v])
        {
            flag=1;
            if (a[v])
            {
                dfs(v,cnt+1);
            }
            else
            {
                dfs(v,0);
            }
        }
    }
    if(!flag)
    {
        ans++;
    }
}

int main()
{

    scanf("%d %d",&n,&m);
    for(i=1; i<=n; i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0; i<n-1; i++)
    {
        int x,y;
        scanf("%d %d",&x,&y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    ans=0;
    dfs(1,a[1]);
    printf("%d\n",ans);
}