Problem: ICP city has an express company whose trucks run from the crossing S to the crossing T. The president of the company is feeling upset because all the roads in the city are one-way, and are severely congested. So, he planned to improve the maximum flow (edge disjoint paths) from the crossing S to the crossing T by reversing the traffic direction on some of the roads.

Your task is writing a program to calculate the maximized flow from S to T by reversing some roads, and the list of the reversed roads.



Input

The first line of a data set contains two integers N (2 \leq N \leq 300) and M (0 \leq M \leq {\rm min} (1\,000,\ N(N-1)/2)). N is the number of crossings in the city and M is the number of roads.

The following M lines describe one-way roads in the city. The i-th line (1-based) contains two integers X_i and Y_i (1 \leq X_i, Y_i \leq N, X_i \neq Y_i). X_i is the ID number (1-based) of the starting point of the i-th road and Y_i is that of the terminal point. The last line contains two integers S and T (1 \leq S, T \leq N, S \neq T, 1-based).

The capacity of each road is 1. You can assume that i \neq j implies either X_i \neq X_j or Y_i \neq Y_j, and either X_i \neq Y_j or X_j \neq Y_i.

Output

In the first line, print the maximized flow by reversing some roads. In the second line, print the number R of the reversed roads. In each of the following R lines, print the ID number (1-based) of a reversed road. You may not print the same ID number more than once.

If there are multiple answers which would give us the same flow capacity, you can print any of them.

Examples

Input

2 1
2 1
2 1


Output

1
0


Input

2 1
1 2
2 1


Output

1
1
1


Input

3 3
3 2
1 2
3 1
1 3


Output

2
2
1
3

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;
template<class S,class T> ostream& operator << (ostream& out,const pair<S,T>& o){ out << "(" << o.first << "," << o.second << ")"; return out; }
template<class T> ostream& operator << (ostream& out,const vector<T> V){ for(int i = 0; i < V.size(); i++){ out << V[i]; if(i!=V.size()-1) out << " ";} return out; }
template<class T> ostream& operator << (ostream& out,const vector<vector<T> > Mat){ for(int i = 0; i < Mat.size(); i++) { if(i != 0) out << endl; out << Mat[i];} return out; }
template<class S,class T> ostream& operator << (ostream& out,const map<S,T> mp){ out << "{ "; for(auto it = mp.begin(); it != mp.end(); it++){ out << it->first << ":" << it->second; if(mp.size()-1 != distance(mp.begin(),it)) out << ", "; } out << " }"; return out; }

/*
 <url:http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2304>
 問題文============================================================
 =================================================================
 解説=============================================================
 ================================================================
 */

struct DINIC{
#define MAX_V 334
    typedef long long ll;
    typedef ll CapType;
    struct edge {
        int to; // 行き先
        CapType cap; // 容量
        int rev; // 逆辺
        edge() {}
        edge(int to, CapType cap, int rev) :to(to), cap(cap), rev(rev) {}
    };
    vector<edge> G[MAX_V]; // グラフの隣接リスト表現
    ll level[MAX_V]; // sからの距離
    ll iter[MAX_V]; // どこまで調べ終わったか
    
    // fromからtoへ向かう容量capの辺をグラフに追加する
    void add_directed_edge(int from, int to, CapType cap) {
        G[from].push_back(edge(to, cap, (int)G[to].size()));
        G[to].push_back(edge(from, 0, (int)G[from].size() - 1));
    }
    void add_undirected_edge(int from, int to, CapType cap) {
        G[from].push_back(edge(to, cap, (int)G[to].size()));
        G[to].push_back(edge(from, cap, (int)G[from].size() - 1));
    }
    
    // sからの最短距離をBFSで計算する
    void bfs(int s){
        fill(level,level+MAX_V,-1);
        queue<int> q;
        level[s] = 0;
        q.push(s);
        while(!q.empty()){
            int v = q.front(); q.pop();
            for(int i = 0; i < (int)G[v].size();i++){
                edge& e = G[v][i];
                if(e.cap > 0 && level[e.to] < 0){
                    level[e.to] = level[v] + 1;
                    q.push(e.to);
                }
            }
        }
    }
    
    // 増加パスをDFSで探す
    CapType dfs(int v, int t,CapType f){
        if(v == t) return f;
        for(ll &i = iter[v]; i < G[v].size();i++){
            edge &e = G[v][i];
            if(e.cap > 0 && level[v] < level[e.to]){
                CapType d = dfs(e.to,t,min(f,e.cap));
                if(d > 0){
                    e.cap -= d;
                    G[e.to][e.rev].cap += d;
                    return d;
                }
            }
        }
        return 0;
    }
    
    CapType max_flow(int s,int t){
        CapType flow = 0;
        for(;;){
            bfs(s);
            if(level[t] < 0) return flow;
            fill(iter,iter+MAX_V,0);
            CapType f;
            while((f = dfs(s,t,LINF)) > 0){
                flow += f;
            }
        }
    }
};

//int main(void) {
//    cin.tie(0); ios::sync_with_stdio(false);
//    int V,E; cin >> V >> E;
//    DINIC Flow;
//    for(int i = 0; i < E;i++){
//        int u,v; ll cap; cin >> u >> v >> cap;
//        Flow.add_directed_edge(u,v,cap);
//    }
//    cout << Flow.max_flow(0, V-1) << endl;
//    return 0;
//}

void solve(){
    int N,M; cin >> N >> M;
    vector<vector<pii>> Graph(N,vector<pii>(N,make_pair(0,-1)));
    DINIC dinic;
    for(int i = 0; i < M;i++){
        int u,v; cin >> u >> v; u--; v--;
        Graph[u][v] = {1,i};
        Graph[v][u] = {-1,i};
        dinic.add_undirected_edge(u,v,1);
    }
    int S,T; cin >> S >> T;
    S--; T--;
    cout << dinic.max_flow(S, T) << endl;
    vector<int> ans;
    for(int i = 0; i < N; i++){
        for(auto e:dinic.G[i]){
            if(e.cap != 0 ) continue;
            int u = i,v = e.to;
            if(Graph[u][v].first == -1){
                ans.push_back(Graph[u][v].second);
            }
        }
    }
    sort(ans.begin(),ans.end());
    cout << ans.size() << endl;
    for(auto A:ans){
        cout << A+1 << endl;
    }
}
int main(void) {
    cin.tie(0); ios_base::sync_with_stdio(false);
    solve();
    return 0;
}