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 <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <utility>
#include <iomanip>
#include <algorithm>
#include <functional>
using namespace std;

typedef long long ll;
typedef vector<int> vint;
typedef vector<long long> vll;
typedef pair<int,int> pint;
typedef pair<long long, long long> pll;

#define MP make_pair
#define PB push_back
#define ALL(s) (s).begin(),(s).end()
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P) 
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P) 
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s << endl; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P) 
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s << endl; }



const int MAX_V = 1000;
const int INF = 1<<29;

struct Edge {
    int rev, from, to, cap, icap;
    Edge(int r, int f, int t, int c) : rev(r), from(f), to(t), cap(c), icap(c) {}
    friend ostream& operator << (ostream& s, const Edge& E) {
        if (E.cap > 0) return s << E.from << "->" << E.to << '(' << E.cap << ')';
        else return s;
    }
};

struct Graph {
    int V;
    vector<Edge> list[MAX_V];
    
    Graph(int n = 0) : V(n) {for (int i = 0; i < MAX_V; ++i) list[i].clear();}
    void init(int n = 0) {V = n; for (int i = 0; i < MAX_V; ++i) list[i].clear();}
	void resize(int n = 0) {V = n;}
    void reset() {for (int i = 0; i < V; ++i) for (int j = 0; j < list[i].size(); ++j) list[i][j].cap = list[i][j].icap;}
    inline vector<Edge>& operator [] (int i) {return list[i];}
    
    Edge &redge(Edge e) {
        if (e.from != e.to) return list[e.to][e.rev];
        else return list[e.to][e.rev+1];
    }
    
    void addedge(int from, int to, int cap) {
        list[from].push_back(Edge(list[to].size(), from, to, cap));
        list[to].push_back(Edge(list[from].size()-1, to, from, 0));
    }
    
    void addbiedge(int from, int to, int cap) {
        list[from].push_back(Edge(list[to].size(), from, to, cap));
        list[to].push_back(Edge(list[from].size()-1, to, from, cap));
    }
    
    friend ostream& operator << (ostream& s, const Graph& G) {
        for (int i = 0; i < G.V; ++i) {s << i << " : " << G.list[i];}
        return s;
    }
};

Graph G;


int level[MAX_V];
int iter[MAX_V];

void dibfs(Graph &G, int s) {
    memset(level, -1, sizeof(level));
    level[s] = 0;
    queue<int> que;
    que.push(s);
    while (!que.empty()) {
        int v = que.front();
        que.pop();
        for (int i = 0; i < G[v].size(); ++i) {
            Edge &e = G[v][i];
            if (level[e.to] < 0 && e.cap > 0) {
                level[e.to] = level[v] + 1;
                que.push(e.to);
            }
        }
    }
}

int didfs(Graph &G, int v, int t, int f) {
    if (v == t) return f;
    for(int &i = iter[v]; i < G[v].size(); ++i) {
        Edge &e = G[v][i], &re = G.redge(e);
        if (level[v] < level[e.to] && e.cap > 0) {
            int d = didfs(G, e.to, t, min(f, e.cap));
            if (d > 0) {
                e.cap -= d;
                re.cap += d;
                return d;
            }
        }
    }
    return 0;
}

int Dinic(Graph &G, int s, int t) {
    int res = 0;
    while (true) {
        dibfs(G, s);
        if (level[t] < 0) return res;
        memset(iter, 0, sizeof(iter));
        int flow;
        while ((flow = didfs(G, s, t, INF)) > 0) {
            res += flow;
        }
    }
}
    
    
int V, E, s, t, a, b;

map<pint,int> ma;



int main() {
    while (cin >> V >> E) {
        G.init(V); ma.clear();
        for (int i = 0; i < E; ++i) {
            cin >> a >> b;
            --a; --b;
            G.addbiedge(a, b, 1);
            ma[ pint(a,b) ] = i+1;
        }
        cin >> s >> t;
        --s; --t;
        
        int res = Dinic(G, s, t);
        
        cout << res << endl;
        vint ans;
        for (int i = 0; i < V; ++i) {
            for (int j = 0; j < G[i].size(); ++j) {
                if (ma.count(pint(i,G[i][j].to)) && G[i][j].cap == 2) {
                    ans.PB(ma[pint(i,G[i][j].to)]);
                    //cout << ma[pint(i,G[i][j].to)] << endl;
                }
            }
        }
        cout << ans.size() << endl;
        for (int i = 0; i < ans.size(); ++i) cout << ans[i] << endl;
    }
    
    return 0;
}