CF796D Police Stations 思维

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.

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

Copy

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

Output

Copy

1
5

Input

Copy

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

Output

Copy

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.

将每个警察局压入queue中，然后 bfs搜索；

如果某个点的 to 已经被访问了，但该条边还没有被访问，ans++;

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/

int n, k, d;
struct node {
    int to;
    int id;
    node(int to,int id):to(to),id(id){}
    node(){}
}edge[maxn];
queue<int>q;
int  ans;
int vis[maxn];
vector<node>vc[maxn];
int fgedge[maxn];

void bfs() {
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = 0; i < vc[u].size(); i++) {
            node tmp = vc[u][i];
            if (fgedge[tmp.id])continue;
            if (vis[tmp.to]) {
                ans++;
                fgedge[tmp.id] = 2;
                continue;
            }
            vis[tmp.to] = 1; fgedge[tmp.id] = 1;
            q.push(tmp.to);
        }
    }
}

int main() {
    //ios::sync_with_stdio(0);
    cin >> n >> k >> d;
    for (int i = 1; i <= k; i++) {
        int x; rdint(x); vis[x] = 1;
        q.push(x);
    }
    for (int i = 1; i < n; i++) {
        int u, v; rdint(u); rdint(v);
        vc[u].push_back(node(v, i));
        vc[v].push_back(node(u, i));
    }
    bfs();
    cout << ans << endl;
    for (int i = 1; i < n; i++) {
        if (fgedge[i] == 2)cout << i << ' ';
    }

    return 0;
}