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

n, m = map(int, input().split())
edge = [list(map(int, input().split())) for _ in range(m)]
lst = [set() for _ in range(n+1)]
for a, b in edge:
    lst[a].add(b)
    lst[b].add(a)

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

check = [0] * (n+1)
q = [1]
next_q = []
tree = [set() for _ in range(n+1)]
top = [1]
while q:
    for oya in q:
        check[oya] = 1
        for ko in lst[oya]:
            if check[ko] == 0:
                check[ko] = 1
                top.append(ko)
                next_q.append(ko)
                tree[oya].add(ko)
                tree[ko].add(oya)
    q = next_q.copy()
    next_q = []

ecount = [0] * (n+1)
for a, b in edge:
    if a in tree[b]:
        continue
    print(a, b)
    ecount[a] += 1

for i in top[::-1][:-1]:
    j = list(tree[i])[0]
    if ecount[i]%2 == 0:
        print(j, i)
        ecount[j] += 1
    else:
        print(i, j)
        ecount[i] += 1
    tree[i].remove(j)
    tree[j].remove(i)