先用trajan缩环变成了一棵树
然后删除了一条边就将树分成了两个部分，一个是删除的那边下面的子树，一个是剩余部分。那么要查询的是两个部分中最大的点的值，和不大于它的最小的点的值。
这样一想就有点像树链剖分啊，树形DP一样求出一颗子树的某个最大值。 又想到是分离出一颗子树，那么就是想到一些dfs序可以对子树进行区间求值。
那么就想到可以预处理出dfs序列，将树的值转化为一个区间值。去掉一颗子树，就是将区间分成三段（可能是两段）。
|____（A）____|____子树（B）____|___（C）____|
求出A,C段最大值中小的那个，和B中最大值。求出最大值。就是答案中的
那么
至于区间最大值，因为不会修改就用RMQ求~树上每个节点的最大值就是连通块里的最大节点值。
// whn6325689
// Mr.Phoebe
// http://blog.csdn.net/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
#define speed std::ios::sync_with_stdio(false);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;
#define CLR(x,y) memset(x,y,sizeof(x))
#define CPY(x,y) memcpy(x,y,sizeof(x))
#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
#define debug(a) cout << #a" = " << (a) << endl;
#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))
#define MID(x,y) (x+((y-x)>>1))
#define ls (idx<<1)
#define rs (idx<<1|1)
#define lson ls,l,mid
#define rson rs,mid+1,r
template<class T>
inline bool read(T &n)
{
T x = 0, tmp = 1;
char c = getchar();
while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
if(c == EOF) return false;
if(c == '-') c = getchar(), tmp = -1;
while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
n = x*tmp;
return true;
}
template <class T>
inline void write(T n)
{
if(n < 0)
{
putchar('-');
n = -n;
}
int len = 0,data[20];
while(n)
{
data[len++] = n%10;
n /= 10;
}
if(!len) data[len++] = 0;
while(len--) putchar(data[len]+48);
}
//-----------------------------------
const int MAXN = 2e5 + 5;
int head_Tu[MAXN], si_Tu;
int head_G[MAXN], si_G;
int par[MAXN], rmax[MAXN];
int pre[MAXN], dfs_clock;
int ans_u[MAXN], ans_v[MAXN];
int in[MAXN], out[MAXN];
int A[MAXN], dmax[MAXN][30];
int n, m;
struct edge
{
int to, next;
}Tu[MAXN<<1], G[MAXN<<1];
struct Node
{
int u, v;
} E[MAXN];
void add_Tu(int st, int ed)
{
Tu[si_Tu].to = ed;
Tu[si_Tu].next = head_Tu[st];
head_Tu[st] = si_Tu++;
}
void add_G(int st, int ed)
{
G[si_G].to = ed;
G[si_G].next = head_G[st];
head_G[st] = si_G++;
}
void init(int _n)
{
for (int i = 0; i <= _n; i++)
par[i] = rmax[i] = i;
}
int Find(int x)
{
if (x == par[x]) return x;
return par[x] = Find(par[x]);
}
void unite(int x, int y)
{
x = Find(x);
y = Find(y);
if (x == y) return ;
par[x] = y;
rmax[y] = max(rmax[x], rmax[y]);
}
int Tarjan(int u, int fa)
{
int lowu = pre[u] = ++dfs_clock;
for (int i = head_Tu[u]; i != -1; i = Tu[i].next)
{
int v = Tu[i].to;
if (v == fa) continue;
if (pre[v] == 0)
{
int lowv = Tarjan(v, u);
lowu = min(lowu, lowv);
if (lowv <= pre[u]) unite(u, v);
}
else lowu = min(lowu, pre[v]);
}
return lowu;
}
void find_bridge(int _n)
{
CLR(pre, 0); dfs_clock = 0;
for (int i = 1; i <= n; i++)
if (!pre[i]) Tarjan(i, -1);
}
void dfs(int u, int fa)
{
in[u] = dfs_clock++;
A[in[u]] = rmax[u];
for (int i = head_G[u]; i != -1; i = G[i].next)
{
int v = G[i].to;
if (v == fa) continue;
dfs(v, u);
}
out[u] = dfs_clock-1;
}
void RMQ_init(int _n)
{
int N = _n;
for (int i = 0; i < N; i++) dmax[i][0] = A[i];
for (int j = 1; (1<<j) <= N; j++)
for (int i = 0; i + (1<<j) - 1 < N; i++)
dmax[i][j] = max(dmax[i][j-1], dmax[i + (1<<(j-1))][j-1]);
}
int RMQ(int L, int R)
{
int k = 0;
while((1<<(k+1)) <= R-L+1) k++;
return max(dmax[L][k], dmax[R-(1<<k)+1][k]);
}
int main()
{
//freopen("input.txt", "r", stdin);
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
CLR(head_Tu, -1); si_Tu = 0;
CLR(head_G, -1); si_G = 0;
CLR(ans_u, -1); CLR(ans_v, -1);
CLR(in, -1); CLR(out, -1);
init(n);
for (int i = 0; i < m; i++)
{
scanf("%d%d", &E[i].u, &E[i].v);
add_Tu(E[i].u, E[i].v);
add_Tu(E[i].v, E[i].u);
}
find_bridge(n);
for (int i = 0; i < m; i++)
{
int u = Find(E[i].u);
int v = Find(E[i].v);
if (u == v)
{
ans_u[i] = ans_v[i] = 0;
}
else
add_G(u, v),add_G(v, u);
}
int root = Find(1);
dfs_clock = 0;
dfs(root, -1);
int len = dfs_clock;
RMQ_init(len);
//debug(si_G);
for (int i = 0; i < m; i++)
{
int u = Find(E[i].u);
int v = Find(E[i].v);
if (in[u] < in[v]) swap(u, v);
if (u != v)
{
//if(i==8){ cout<<" "; debug(in[u]);debug(out[u]);}
int x = RMQ(in[u], out[u]);
int y = 0;
if (in[u] > 0) y = max(y, RMQ(0, in[u]-1));
if (out[u]+1 < len) y = max(y, RMQ(out[u]+1, len-1));
//if(i==8){cout<<x<<" "<<y<<endl;}
ans_u[i] = min(x, y);
ans_v[i] = ans_u[i] + 1;
}
}
for (int i = 0; i < m; i++)
printf("%d %d\n", ans_u[i], ans_v[i]);
}
return 0;
}