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 PYTHON3):

Solution:
from __future__ import division, print_function

import os
import sys
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

# region fastio
# Credits
# # template credits to cheran-senthil's github Repo

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

def main():
    from types import GeneratorType
    def bootstrap(f, stack=[]):
        def wrappedfunc(*args, **kwargs):
            if stack:
                return f(*args, **kwargs)
            else:
                to = f(*args, **kwargs)
                while True:
                    if type(to) is GeneratorType:
                        stack.append(to)
                        to = next(to)
                    else:
                        stack.pop()
                        if not stack:
                            break
                        to = stack[-1].send(to)
                return to
        return wrappedfunc
    import time
    class Graph(object):

        def __init__(self):
            self.neighbours = {}
        
        def __repr__(self):
            s = ''
            for i in self.neighbours:
                s += str(i)+': '+' '.join(map(str,self.neighbours[i])) + '\n'
            return s
        
        def add_node(self, node):
            self.neighbours[node] = set()
        
        def add_edge(self, edge):
            u, v = edge
            self.neighbours[u].add(v)
            self.neighbours[v].add(u)

        @bootstrap
        def dfs(self, node, par):
            visited[node] = 1
            intime[node] = lowtime[node] = timer[0]
            timer[0] += 1
            for child in self.neighbours[node]:
                # if cur node's child is parent than for forward edge it is valid ancestor
                if child == par:
                    continue
                # even after checking for parent it is already visited than it is a back edge
                if visited[child] == 1:
                    if intime[node] > intime[child]:
                        lis.append([node,child])
                    # This is a back edge
                    # So we are updating the lowtime of cur node with intime of 
                    lowtime[node] = min(lowtime[node],intime[child])
                else:
                    # This is a forward edge
                    lis.append([node,child])
                    yield self.dfs(child, node)
                    # if low[child]<in[node]
                    # even if you remove this edge child is still connected from another path. so can't be a bridge
                    if lowtime[child] > intime[node]:
                        flag[0] = 0
                    lowtime[node] = min(lowtime[node],lowtime[child])
            yield
                

        
        def bfs(self, node):
            q = queue()
            visited[node] = 1
            q.append(node)
            while q:
                cur = q.popleft()
                for child in self.neighbours[cur]:
                    if visited[child] == 0:
                        q.append(child)
                        visited[child] = 1

    g = Graph()
    n,m=map(int,input().split())
    for i in range(n):
        g.add_node(i+1)
    for i in range(m):
        x, y = map(int, input().split())
        g.add_edge((x, y))
    visited = [0] * (n + 1)
    intime = [0] * (n + 1)
    lowtime = [0] * (n + 1)
    timer = [0]
    flag = [1]
    lis = []
    g.dfs(1, -1)
    if flag[0]:
        for i in lis:
            print(' '.join(map(str,i)))
    else:
        print(0)


if __name__ == "__main__":
    main()