#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10, M = 4e6 + 10;
typedef pair<int, int> PII;
#define x first
#define y second

// 链式前向星
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c = 0) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}

int dfn[N], low[N], ts, root;
set<PII> _set;

void tarjan(int u, int fa) {
    dfn[u] = low[u] = ++ts;

    // 对比有向图强连通分量的代码，这里没有入栈的操作，原因是本题只想求割边，并不是想缩点，
    // 怀疑缩点时就需要加入stk,in_stk等操作了，待验证
    for (int i = h[u]; ~i; i = ne[i]) {
        int v = e[i];
        if (v == fa) continue;
        if (!dfn[v]) {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            // 记录割边,论证过的结论
            if (low[v] > dfn[u]) _set.insert({u, v}); // SET天生会排序，按first由小到大
        } else
            low[u] = min(low[u], dfn[v]);
    }
}

int n, m;
int main() {
#ifndef ONLINE_JUDGE
    freopen("P1656.in", "r", stdin);
#endif
    memset(h, -1, sizeof h);
    scanf("%d %d", &n, &m);
    while (m--) {
        int a, b;
        scanf("%d %d", &a, &b);
        if (a != b) add(a, b), add(b, a);
    }

    for (root = 1; root <= n; root++)
        if (!dfn[root]) tarjan(root, -1);

    for (auto i : _set) printf("%d %d\n", i.x, i.y);
    return 0;
}