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:
def MI():
    return map(int,input().split())
def I():
    return int(input())
def LI():
    return [int(i) for i in input().split()]

n,m=MI()

g=[[] for i in range(n)]
v=[0]*n
f=[0]*n
e=[]

for i in range(m):
    a,b=map(int,input().split())
    a,b=a-1,b-1
    g[a].append(b)
    g[b].append(a)
    e.append((a,b))

def get_tree(st):
    p,q=[],[]
    q.append(st)
    v[st]=1
    
    while q:
        cur=q.pop()
        for nb in g[cur]:
            if v[nb]!=1:
                q.append(nb)
                p.append((cur,nb))
                v[nb]=1
    return p

def solve():
    tr=get_tree(0)
    st=set(tr)
    rt=[]
    for a,b in e:
        if (a,b) in st or (b,a) in st:
            pass
        else:
            rt.append((a,b))
            f[a]=1-f[a]
    for a,b in tr[::-1]:
        if f[b]==0:
            rt.append((a,b))
            f[a]=1-f[a]
        else:
            rt.append((b,a))
            f[b]=1-f[b]
    return rt

if m%2==1:
    print(-1)
else:
    for a,b in solve():
        print(a+1,b+1)