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:
# -*- coding: utf-8 -*-
"""
Created on Thu Jul 18 12:05:59 2019

@author: ctiuser
"""
import sys
sys.setrecursionlimit(10**6)

N, M = map(int,input().split())
A = [list(map(int,input().split())) for i in range(M)]

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

class UnionFind():
    def __init__(self,size):
        self.table = [-1 for i in range(size+1)]
    def find(self,x):
        if self.table[x] < 0:
            return x
        else:
            return self.find(self.table[x])
    def union(self,x,y):
        s1 = self.find(x)
        s2 = self.find(y)
        if s1 == s2:
            return False
        if self.table[s1] < self.table[s2]:
            self.table[s1] = self.table[s1] + self.table[s2]
            self.table[s2] = self.find(x)
        else:
            self.table[s2] = self.table[s1] + self.table[s2]
            self.table[s1] = self.find(y)
        return True

UF = UnionFind(N)
ans = []
T = [[-1] for i in range(N+1)]
color = [-1 for i in range(N+1)]
counta = [0 for i in range(N+1)]
B = []
C = [[] for i in range(N+1)]
for c in A:
    if UF.union(c[0],c[1]):
        B.append(c)
        C[c[0]].append(c[1])
        C[c[1]].append(c[0])
    else:
        ans.append(c)
        counta[c[0]] += 1

def b(x):
    color[x] = 1
    for c in C[x]:
        if color[c] == -1:
            T[c][0] = x
            T[x].append(c)
            b(c)
b(1)

edge_leaf = []
for i in range(1,N+1):
    if len(T[i]) == 1:
        edge_leaf.append(i)

def a(x):
    p = T[x][0]
    if p == -1:
        return 0
    if counta[x] % 2 == 0:
        ans.append([p,x])
        counta[p] += 1
    else:
        ans.append([x,p])
    return p

for c in edge_leaf:
    point = c
    while color[point] == len(T[point]):
        point = a(point)
        color[point] += 1


for c in ans:
    print(' '.join(map(str,c)))