Codeforces Round #408 (Div. 2) D.Police Stations【Bfs+思维】

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本文介绍了一种在保证每个普通节点到特殊节点距离不超过给定阈值的前提下，最大化裁减树形结构中边数量的算法。通过广度优先搜索策略确保了特殊节点覆盖的有效性和最优化，最终实现了道路维护成本的最小化。

Inzane finally found Zane with a lot of money to spare, so they together decided to establish a country of their own.

Ruling a country is not an easy job. Thieves and terrorists are always ready to ruin the country's peace. To fight back, Zane and Inzane have enacted a very effective law: from each city it must be possible to reach a police station by traveling at most d kilometers along the roads.

$\text{[math]}$

There are n cities in the country, numbered from 1 to n, connected only by exactly n - 1 roads. All roads are 1 kilometer long. It is initially possible to travel from a city to any other city using these roads. The country also has k police stations located in some cities. In particular, the city's structure satisfies the requirement enforced by the previously mentioned law. Also note that there can be multiple police stations in one city.

However, Zane feels like having as many as n - 1 roads is unnecessary. The country is having financial issues, so it wants to minimize the road maintenance cost by shutting down as many roads as possible.

Help Zane find the maximum number of roads that can be shut down without breaking the law. Also, help him determine such roads.

Input

The first line contains three integers n, k, and d (2 ≤ n ≤ 3·10⁵, 1 ≤ k ≤ 3·10⁵, 0 ≤ d ≤ n - 1) — the number of cities, the number of police stations, and the distance limitation in kilometers, respectively.

The second line contains k integers p₁, p₂, ..., p_k (1 ≤ p_i ≤ n) — each denoting the city each police station is located in.

The i-th of the following n - 1 lines contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i) — the cities directly connected by the road with index i.

It is guaranteed that it is possible to travel from one city to any other city using only the roads. Also, it is possible from any city to reach a police station within d kilometers.

Output

In the first line, print one integer s that denotes the maximum number of roads that can be shut down.

In the second line, print s distinct integers, the indices of such roads, in any order.

If there are multiple answers, print any of them.

Examples

Input

Output

1
5

Input

Output

2
4 5

Note

In the first sample, if you shut down road 5, all cities can still reach a police station within k = 4 kilometers.

In the second sample, although this is the only largest valid set of roads that can be shut down, you can print either 4 5 or 5 4 in the second line.

题目大意：

给你一个树，其中有N个点，N-1条无向边，现在有M个特殊点，问你最多可以切割多少条边，使得每个普通点到最近的特殊点的距离小于等于d.

保证有解。

思路：

保证有解，其实这个d就没有卵用了。

我们只要将这M个特殊点放到队列中，此时Bfs肯定有一个时间优先度。

那么我们直接Bfs.使得每次队头向下走一步，走到没有走过的点上边，对应记录每条边是否使用过即可。

这样去分块一定是最优的。

剩余没用过的边一定是最多的任意切割的边。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
using namespace std;
struct node
{
    int to,pos;
}now,nex;
vector<node >mp[300800];
int vis[300800];
int ans[300800];
int main()
{
    int n,m,k;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        queue<int >s;
        memset(vis,0,sizeof(vis));
        memset(ans,0,sizeof(ans));
        for(int i=1;i<=n;i++)mp[i].clear();
        for(int i=0;i<m;i++)
        {
            int x;
            scanf("%d",&x);
            s.push(x);
            vis[x]=1;
        }
        for(int i=1;i<=n-1;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            now.to=y;
            now.pos=i;
            mp[x].push_back(now);
            now.to=x;
            now.pos=i;
            mp[y].push_back(now);
        }
        while(!s.empty())
        {
            int u=s.front();
            s.pop();
            for(int i=0;i<mp[u].size();i++)
            {
                int v=mp[u][i].to;
                if(vis[v]==0)
                {
                    ans[mp[u][i].pos]=1;
                    vis[v]=1;
                    s.push(v);
                }
            }
        }
        int cnt=0;
        for(int i=1;i<=n-1;i++)if(ans[i]==0)cnt++;
        printf("%d\n",cnt);
        for(int i=1;i<=n-1;i++)
        {
            if(ans[i]==0)printf("%d ",i);
        }
        printf("\n");
    }
}