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:
# Euler Tourとimos法で殴る。
from bisect import bisect


N, M = map(int, input().split())
ABs = [tuple(map(int, input().split())) for _ in range(M)]

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

inv_ABs = [dict() for _ in range(N+1)]
for i, (A, B) in enumerate(ABs):
    inv_ABs[A][B] = i
    inv_ABs[B][A] = i

reverse_ABs = [False] * M

iseven = [True] * (N+1)
for A, B in ABs:
    if not iseven[B]: # B -> A
        iseven[B] = True
        reverse_ABs[inv_ABs[A][B]] = True
    else:
        iseven[A] = (not iseven[A])


# Euler Tour
start = 1
stack = [start]
parent = [0] * (N+1)
parent[start] = start
et = []
while stack:
    v = stack.pop()
    if v >= 0:
        et.append(v)
        for nv in inv_ABs[v]:
            if parent[nv] == 0:
                parent[nv] = v
                stack.append(~nv)
                stack.append(nv)
    else:
        et.append(parent[~v])

# print(et)

imos = [0] * (len(et) + 1)
odd_nodes = set((i for i, e in enumerate(iseven) if not e))
for i, v in enumerate(et):
    if v in odd_nodes:
        odd_nodes.remove(v)
        imos[i] += 1
    imos[i+1] += imos[i]

for i in range(len(et) - 1):
    if imos[i] % 2 == 1:
        reverse_ABs[inv_ABs[et[i]][et[i+1]]] = (not reverse_ABs[inv_ABs[et[i]][et[i+1]]] )

for rflg, (A, B) in zip(reverse_ABs, ABs):
    if rflg:
        print(B, A)
    else:
        print(A, B)