Problem: 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.

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.

Examples

Input

4 4
1 2
2 3
3 4
4 1


Output

1 2
1 4
3 2
3 4


Input

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


Output

-1

Solution (in PYTHON3):

Solution:

"""
無理な場合を考えてみよう
→全体の辺の本数が奇数なら絶対無理

それ以外は？
説1:辺が偶数なら絶対作れる説

辺を選択し、そのどちらか片方にのみ xor1する
→すべての辺の数字を0にできるか？？(初期値は0)
なんかできそう

辺の向きを変えることで1を伝播させることが出来る
→同じ点にぶつけられる！！
ので、偶数本なら絶対できる


どうやって構成する？
適当に置く→1の点を適当に2つ選び、パス上にある辺を全部反転させる
確かにできるけど…遅くね？

もっと簡単にできないか？
適当に木を取る→深い方から整合性を合わせていけば終わり(最終的には根に収束するので)
各点に関して動かすのは高々1辺だけだから、木を取った辺のみ考えて反転させる
(それ以外は適当に置いておく)

dfsで終わり→解けた！！！
"""
import sys
sys.setrecursionlimit(200000)
N,M = map(int,input().split())

lisN = [ [] for i in range(N) ]
xorlis = [0] * N
AB = []

if M % 2 == 1:
    print (-1)
    sys.exit()

for i in range(M):

    a,b = map(int,input().split())
    AB.append([a,b])
    a -= 1
    b -= 1

    lisN[a].append([b,i])
    lisN[b].append([a,i])
    xorlis[b] ^= 1

visited = [False] * N
def dfs(now):

    visited[now] = True

    for nex,eind in lisN[now]:
        if visited[nex]:
            continue
        cat = dfs(nex)
        if cat != 0:
            xorlis[nex] ^= 1
            xorlis[now] ^= 1
            t = AB[eind][1]
            AB[eind][1] = AB[eind][0]
            AB[eind][0] = t

    return xorlis[now]

dfs(0)
#print (xorlis)
if 1 in xorlis:
    print (-1)
    sys.exit()

else:
    for i,j in AB:
        print (j,i)