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:
# YouTube
from collections import defaultdict
import sys

sys.setrecursionlimit(10 ** 6)

def dfs(i, oi=-1, d=0):
    dep[d] += [i]
    oya[i] = oi
    for ki in to[i]:
        if ki == oi: continue
        if oya[ki] > -2:
            if i < ki:
                print(i + 1, ki + 1)
                ne[i] += 1
            continue
        dfs(ki, i, d + 1)

to = defaultdict(list)
dep = defaultdict(list)
n, m = map(int, input().split())
if m % 2:
    print(-1)
    exit()
oya = [-2] * n
ne = [0] * n
for _ in range(m):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    to[a] += [b]
    to[b] += [a]
dfs(0)
for k, v in sorted(dep.items(), reverse=True):
    for i in v:
        if ne[i] % 2:
            print(i + 1, oya[i] + 1)
        else:
            print(oya[i] + 1, i + 1)
            ne[oya[i]] += 1
    if k == 1: break