D. Book of Evil time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Paladin Manao caught the trail of the ancient Book of Evil in a swampy area. This area contains n settlements numbered from 1 to n. Moving through the swamp is very difficult, so people tramped exactly n - 1 paths. Each of these paths connects some pair of settlements and is bidirectional. Moreover, it is possible to reach any settlement from any other one by traversing one or several paths.

The distance between two settlements is the minimum number of paths that have to be crossed to get from one settlement to the other one. Manao knows that the Book of Evil has got a damage range d. This means that if the Book of Evil is located in some settlement, its damage (for example, emergence of ghosts and werewolves) affects other settlements at distance d

 or less from the settlement where the Book resides.

Manao has heard of m settlements affected by the Book of Evil. Their numbers are p₁, p₂, ..., p_m. Note that the Book may be affecting other settlements as well, but this has not been detected yet. Manao wants to determine which settlements may contain the Book. Help him with this difficult task.

Input

The first line contains three space-separated integers n, m and d (1 ≤ m ≤ n ≤ 100000; 0 ≤ d ≤ n - 1). The second line contains mdistinct space-separated integers p₁, p₂, ..., p_m (1 ≤ p_i ≤ n). Then n - 1 lines follow, each line describes a path made in the area. A path is described by a pair of space-separated integers a_i and b_i representing the ends of this path.

Output

Print a single number — the number of settlements that may contain the Book of Evil. It is possible that Manao received some controversial information and there is no settlement that may contain the Book. In such case, print 0.

Examples input

6 2 3
1 2
1 5
2 3
3 4
4 5
5 6

output

3

Note

Sample 1. The damage range of the Book of Evil equals 3 and its effects have been noticed in settlements 1 and 2. Thus, it can be in settlements 3, 4 or 5.

題意：就是一棵樹上，有一些點被來自東方的神祕力量影響的，力量影響範圍是d，為可能的力量源有幾個。

思路：相當於是找到距離這m的點的距離都不小於d的點的個數。

先從任意一個點一次dfs，找到m個點中距離最遠的點，標記為max1，在以max1開始，dfs一遍，從陣列d1記錄各個點的距離，找到距離最遠的點max2，再從max2開始跑一遍dfs，用d2記錄距離。遍歷所有點，到max1和max2的距離都小於d的點就成立。

#include<cstdio>
#include<cstring>
#include <string>
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
using namespace std;
#define MAXN 200010
#define INF 1000000000

int d1[MAXN],d2[MAXN];
int h[MAXN];
int head[MAXN];
int p[MAXN];
set <int> s;
vector <int> ans;
struct edge{int v;int next;};
edge G[MAXN*2];
int num=0;
int vis[MAXN];
void add(int u,int v){G[num].v=v;G[num].next=head[u];head[u]=num++;};
void dfs(int u,int d[])
{
    vis[u]=true;
    for(int k=head[u];k!=-1;k=G[k].next){
        int v=G[k].v;
        if(!vis[v]){
            d[v]=d[u]+1;
            dfs(v,d);
        }
    }
}
int n,m,d;
int main(){
    scanf("%d%d%d",&n,&m,&d);
    for(int i=0;i<m;i++){
        scanf("%d",p+i);
        s.insert(p[i]);
    }
    for(int i=1;i<=n;i++){
        if(s.count(i)==0)
            ans.push_back(i);
    }
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    for(int i=0;i<n-1;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        add(a,b);
        add(b,a);
    }
    h[1]=0;
    dfs(1,h);
    int max1=0;
    for(int i=0;i<m;i++){
        if(h[p[i]]>h[p[max1]])
            max1=i;
    }
    memset(vis,0,sizeof(vis));
    d1[p[max1]]=0;
    dfs(p[max1],d1);
    int max2=0;
    for(int i=0;i<m;i++){
        if(d1[p[i]]>d1[p[max2]])
            max2=i;
    }
    memset(vis,0,sizeof(vis));
    d2[p[max2]]=0;
    dfs(p[max2],d2);
    int cnt=0;
    for(int i=1;i<=n;i++){
        if(d1[i]<=d&&d2[i]<=d)
            cnt++;
    }
    printf("%d\n",cnt);
    return 0;
}