#include <bits/stdc++.h>
using namespace std;

const int N = 1010, M = N << 2;

int ts;
int dfn[N], low[N];

// 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 out[M]; // 是不是成对变换输出过
void tarjan(int u, int fa) {
    dfn[u] = low[u] = ++ts;
    for (int i = h[u]; ~i; i = ne[i]) {
        int v = e[i];
        if (v == fa) continue;

        if (out[i]) continue;    // 对边已经输出过，那么，这条反边不能输出。因为如果是割边的话，两条边在u->v时就都已经输出完了
        out[i] = out[i ^ 1] = 1; // 标识成对变换输出过
        printf("%d %d\n", u, v); // 输出u->v,同时，需要检查 v->u是桥的话，还输出v->u

        if (!dfn[v]) {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if (dfn[u] < low[v])         // 割边
                printf("%d %d\n", v, u); // 割边需要输出两条
        } else
            low[u] = min(low[u], dfn[v]);
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("UVA610.in", "r", stdin);
#endif
    int n, m, cas = 0;
    while (scanf("%d%d", &n, &m), n + m) {
        idx = ts = 0;
        memset(h, -1, sizeof h);
        memset(dfn, 0, sizeof(dfn));
        memset(low, 0, sizeof(low)); // Tips:有些人的代码，low也是可以不用清空的
        memset(out, 0, sizeof out);

        while (m--) {
            int a, b;
            scanf("%d%d", &a, &b);
            add(a, b), add(b, a);
        }
        printf("%d\n\n", ++cas);
        for (int i = 1; i <= n; i++)
            if (!dfn[i]) tarjan(i, -1);
        puts("#");
    }
    return 0;
}