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:
import sys
input=sys.stdin.readline
from bisect import bisect_left 
from collections import deque

def count_edge(a,b):
    global edge,AB
    if a>b:
        a,b=b,a
    edge[bisect_left(AB,(a,b))]+=1

def main():
    global edge,AB
    N,M=map(int,input().split())
    data=[[] for i in range(N+1)]
    AB=[]
    edge=[0]*M
    count=[0]*(N+1)
    for i in range(M):
        A,B=map(int,input().split())
        if A>B:
            A,B=B,A
        AB.append((A,B))
        data[A].append(B)
        data[B].append(A)
        count[A]^=1
    AB.sort()

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

    parent=[0]*(N+1)
    que=[1]
    parent[1]=1
    lsls=deque([])
    while que:
        h=[]
        for u in que:
            for v in data[u]:
                if parent[v]==0:
                    parent[v]=u
                    h.append(v)
                    if count[v]%2==1:
                        lsls.appendleft(v)
        que=h

    for i in lsls:
        if count[i]==0:
            continue
        else:
            count[i]=0
            while True:
                j=parent[i]
                if i==j:
                    break
                count_edge(i,j)
                if count[j]==1:
                    count[j]=0
                    break
                i=j
            
    for i in range(M):
        if edge[i]%2==0:
            print(*AB[i])
        else:
            print(*AB[i][::-1])
    
if __name__=="__main__":
    main()