Problem: Bertown has n junctions and m bidirectional roads. We know that one can get from any junction to any other one by the existing roads. 

As there were more and more cars in the city, traffic jams started to pose real problems. To deal with them the government decided to make the traffic one-directional on all the roads, thus easing down the traffic. Your task is to determine whether there is a way to make the traffic one-directional so that there still is the possibility to get from any junction to any other one. If the answer is positive, you should also find one of the possible ways to orient the roads.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 105, n - 1 ≤ m ≤ 3·105) which represent the number of junctions and the roads in the town correspondingly. Then follow m lines, each containing two numbers which describe the roads in the city. Each road is determined by two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers of junctions it connects.

It is guaranteed that one can get from any junction to any other one along the existing bidirectional roads. Each road connects different junctions, there is no more than one road between each pair of junctions.

Output

If there's no solution, print the single number 0. Otherwise, print m lines each containing two integers pi and qi — each road's orientation. That is the traffic flow will move along a one-directional road from junction pi to junction qi. You can print the roads in any order. If there are several solutions to that problem, print any of them.

Examples

Input

6 8
1 2
2 3
1 3
4 5
4 6
5 6
2 4
3 5


Output

1 2
2 3
3 1
4 5
5 6
6 4
4 2
3 5


Input

6 7
1 2
2 3
1 3
4 5
4 6
5 6
2 4


Output

0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const int MaxN = 4e5 + 10;
const int MAXN = 120 * 1000;
void fun() {}
long long phi(long long n) {
  long long result = n;
  for (long long p = 2; p * p <= n; ++p) {
    if (n % p == 0) {
      while (n % p == 0) n /= p;
      result -= result / p;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
long long mult(long long a, long long b, long long mod) {
  long long res = 0;
  a %= mod;
  while (b) {
    if (b & 1) res = (res + a) % mod;
    a = (2 * a) % mod;
    b >>= 1;
    return res;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
vector<long long> factors(long long n) {
  vector<long long> f;
  for (long long x = 2; x * x <= n; x++) {
    if (n % x == 0) {
      f.push_back(x);
      if ((n / x) != x) f.push_back(n / x);
    }
  }
  return f;
}
int cD(long long n) { return floor(log10(n) + 1); }
long long powmod(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long inv(long long x) { return powmod(x, mod - 2); }
long long cat(long long n) {
  long long final = 1;
  for (int i = 0; i < n; i++) {
    final = final * (2 * n - i) % mod * inv(1 + i) % mod;
  }
  final = final * inv(1 + n) % mod;
  return final;
}
long long bin(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long u = bin(x, y / 2);
  if (y % 2 == 0) {
    return u * u % mod;
  } else {
    return u * u % mod * x % mod;
  }
}
long long modFact(long long n, long long mod) {
  if (n >= mod) return 0;
  long long result = 1;
  for (int i = 1; i <= n; i++) result = (result * i) % mod;
  return result;
}
double nthRoot(double n, int k) {
  return pow(k, (1.0 / k) * (log(n) / log(k)));
}
long long MAX = 1e6;
long long n, k, q, c, r, h;
long long l, t;
string s;
std::vector<int> adj[100001];
long long visited[100001];
long long chi[100001];
long long dist[100001];
long long timer;
long long in[100001];
long long low[100001];
long long d;
long long cc;
bool edge = false;
vector<pair<int, int>> el;
void dfs(int curr, int par) {
  visited[curr] = 1;
  low[curr] = in[curr] = timer++;
  for (int child : adj[curr]) {
    if (child == par) continue;
    if (visited[child]) {
      low[curr] = min(low[curr], in[child]);
      if (in[curr] > in[child]) el.push_back({curr, child});
    } else {
      dfs(child, curr);
      if (low[child] > in[curr]) {
        edge = true;
        return;
      }
      el.push_back({curr, child});
      low[curr] = min(low[curr], low[child]);
    }
  }
}
int main() {
  fun();
  ios_base::sync_with_stdio(false);
  long long n, m, e;
  long long t;
  cin >> n >> e;
  for (int i = 0; i < e; i++) {
    cin >> l >> r;
    adj[l].push_back(r);
    adj[r].push_back(l);
  }
  dfs(1, -1);
  if (edge == true) {
    cout << 0 << "\n";
  } else {
    for (int i = 0; i < el.size(); i++)
      cout << el[i].first << " " << el[i].second << "\n";
  }
  return 0;
}