/*=========================================
    无向图求割点和桥
    复杂度：O(E + V)
    割点：
        1：若k为深搜树的根Root，当且仅当k的儿子数（分支数）>=2时k为割点；
        2：若k为搜索树的中间结点（即k既不为根也不为叶），那么k必然有father和son，若low[son]>= dfn[k]，则k必然为割点。
    割桥：
        如果low[v] > dfn[u] 则(u,v)为割桥 (需要判断是否有重边，如果有重边一定不是割边）
=========================================*/
/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 11000;
const int maxm = 110000;
int n, m;
struct node {
    int v, id, tot;
    node(int _v, int _id, int _tot) {
        v = _v, id = _id, tot = _tot;
    }
};
vector<node> g[maxn];
int dfn[maxn], low[maxn], vis[maxn], sub[maxn]; //sub记录每个割点分为多少块
bool cut[maxn];
int bridge[maxm], cnt;
void dfs(int u, int f, int dep) {
    vis[u] = 1;
    dfn[u] = low[u] = dep;
    int child = 0;
    for (int i = 0; i < g[u].size(); i++) if (g[u][i].v != f) {
            int v = g[u][i].v;
            if (vis[v] == 1 && dfn[v] < low[u]) low[u] = dfn[v];    //回边更新low值
            if (vis[v] == 0) {
                dfs(v, u, dep + 1);
                child++;
                if (low[v] < low[u]) low[u] = low[v];  //通过儿子回到祖先
                if (low[v] > dfn[u] && g[u][i].tot == 1) bridge[cnt++] = g[u][i].id;
            }
        }
    vis[u] = 2;
}
void cut_bridge() {
    cnt = 0;
    memset(vis, 0, sizeof(int) * (n + 10));
    dfs(1, -1, 0); //i为一个点
    //如果多个连通块，需要对每一个连通块调用dfs
}
int main () {
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        for (int i = 0; i <= n; i++) g[i].clear();
        int u, v;
        for (int i = 1; i <= m; i++) {
            scanf("%d%d", &u, &v);
            bool flag = false;
            for (int j = 0; j < g[u].size(); j++) if (g[u][j].v == v) {
                g[u][j].tot++;
                flag = true;
                break;
            }
            if (flag) for (int j = 0; j < g[v].size(); j++) if (g[v][j].v == u) {
                g[v][j].tot++;
                break;
            }
            if (!flag)
                g[u].pb(node(v, i, 1)), g[v].pb(node(u, i, 1));
        }
        cut_bridge();
        sort(bridge, bridge + cnt);
        printf("%d\n", cnt);
        rep(i, 0, cnt) printf("%d%c", bridge[i], i == cnt - 1 ? '\n' : ' ');
        if (T > 0) puts("");
    }
    return 0;
}