Problem:
Score : 700 points

Problem Statement

You are given a simple connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the i-th edge connects Vertex A_i and Vertex B_i.
Takahashi will assign one of the two possible directions to each of the edges in the graph to make a directed graph.
Determine if it is possible to make a directed graph with an even number of edges going out from every vertex. If the answer is yes, construct one such graph.

Notes

An undirected graph is said to be simple when it contains no self-loops or multiple edges.

Constraints

2 \leq N \leq 10^5

N-1 \leq M \leq 10^5

1 \leq A_i,B_i \leq N (1\leq i\leq M)

The given graph is simple and connected.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_M B_M

Output

If it is impossible to assign directions to satisfy the requirement, print -1.
Otherwise, print an assignment of directions that satisfies the requirement, in the following format:

C_1 D_1
:
C_M D_M

Here each pair (C_i, D_i) means that there is an edge directed from Vertex C_i to Vertex D_i. The edges may be printed in any order.

Sample Input 1

4 4
1 2
2 3
3 4
4 1

Sample Output 1

1 2
1 4
3 2
3 4

After this assignment of directions, Vertex 1 and 3 will each have two outgoing edges, and Vertex 2 and 4 will each have zero outgoing edges.

Sample Input 2

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

Sample Output 2

-1

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

#!/usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
import itertools
sys.setrecursionlimit(10**5)
stdin = sys.stdin
bisect_left = bisect.bisect_left
bisect_right = bisect.bisect_right
def LI(): return list(map(int, stdin.readline().split()))
def LF(): return list(map(float, stdin.readline().split()))
def LI_(): return list(map(lambda x: int(x)-1, stdin.readline().split()))
def II(): return int(stdin.readline())
def IF(): return float(stdin.readline())
def LS(): return list(map(list, stdin.readline().split()))
def S(): return list(stdin.readline().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')

#A
def A():
    n = II()
    a = LI()
    a.sort()
    x = list(set(a))
    if len(x) > 3:
        print("No")
    elif len(x) == 3:
        x.sort()
        x1 = bisect_right(a, x[0])
        x2 = bisect_right(a, x[1]) - bisect_left(a, x[1])
        x3 = bisect_right(a, x[2]) - bisect_left(a, x[2])
        if ((x[0] ^ x[1] ^ x[2]) == 0) and (x1 == x2 == x3):
            print("Yes")
        else:
            print("No")
    elif len(x) == 2:
        x1 = bisect_right(a, 0)
        x2 = bisect_right(a, x[1]) - bisect_left(a, x[1])
        if x1 * 2 == x2:
            print("Yes")
        else:
            print("No")
    elif len(x) == 1:
        if x[0] == 0:
            print("Yes")
        else:
            print("No")
    return

#B
def B():
    n, m = LI()
    if m % 2:
        print(-1)
        return
    edg = [[] for _ in range(n)]
    ins = defaultdict(int)
    dag = defaultdict(list)
    for _ in range(m):
        a, b = LI_()
        edg[a].append(b)
        edg[b].append(a)
        dag[a].append(b)
        ins[a] += 1
    for i in range(n):
        if ins[i] % 2:
            q = deque()
            q.append((i, -1 ,[i]))
            while ins[i] % 2:
                now, pre, route_ = q.popleft()
                for k in edg[now]:
                    if pre == k:
                        continue
                    route = route_[::1]
                    route.append(k)
                    if ins[k] % 2:
                        for l in range(len(route)-1):
                            a = route[l]
                            b = route[l + 1]
                            if b in dag[a]:
                                dag[a].remove(b)
                                dag[b].append(a)
                            elif a in dag[b]:
                                dag[b].remove(a)
                                dag[a].append(b)
                        ins[i] -= 1
                        ins[k] += 1
                        break
                    else:
                        q.append((k, now, route))
    for e1,el2 in dag.items():
        for e2 in el2:
            print(e1+1,e2+1)
    return

#C
def C():
    return

#D
def D():
    return

#E
def E():
    return

#F
def F():
    return

#G
def G():
    return

#H
def H():
    return

#Solve
if __name__ == '__main__':
    B()