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#include <bits/stdc++.h>
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using namespace std;
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const int N = 5e5 + 10, M = 4e6 + 10;
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typedef pair<int, int> PII;
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#define x first
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#define y second
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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int dfn[N], low[N], ts, root;
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set<PII> _set;
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void tarjan(int u, int fa) {
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dfn[u] = low[u] = ++ts;
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// 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，
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// 怀疑缩点时就需要加入stk,in_stk等操作了，待验证
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (v == fa) continue;
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if (!dfn[v]) {
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tarjan(v, u);
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low[u] = min(low[u], low[v]);
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// 记录割边,论证过的结论
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if (low[v] > dfn[u]) _set.insert({u, v}); // SET天生会排序，按first由小到大
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} else
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low[u] = min(low[u], dfn[v]);
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}
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}
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int n, m;
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("P1656.in", "r", stdin);
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#endif
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memset(h, -1, sizeof h);
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scanf("%d %d", &n, &m);
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while (m--) {
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int a, b;
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scanf("%d %d", &a, &b);
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if (a != b) add(a, b), add(b, a);
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}
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for (root = 1; root <= n; root++)
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if (!dfn[root]) tarjan(root, -1);
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for (auto i : _set) printf("%d %d\n", i.x, i.y);
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return 0;
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}
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