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 os
import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7

N, M = list(map(int, sys.stdin.readline().split()))
AB = [list(map(int, sys.stdin.readline().split())) for _ in range(M)]

graph = [[] for _ in range(N + 1)]
for a, b in AB:
    graph[a].append(b)
    graph[b].append(a)


def spanning_tree(root, ret=None, seen=None):
    if ret is None:
        ret = [set() for _ in range(N + 1)]
    if seen is None:
        seen = [False] * (N + 1)
        seen[root] = True
    for v in graph[root]:
        if seen[v]:
            continue
        ret[root].add(v)
        seen[v] = True
        ret = spanning_tree(v, ret, seen)
    return ret


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

# 全域木
root = 1
st = spanning_tree(root)

# 出次数
outdeg = [0] * (N + 1)
ans = []
# 使わなかった辺に向きをつける
for a, b in AB:
    if a not in st[b] and b not in st[a]:
        ans.append((a, b))
        outdeg[a] += 1


def dfs(par):
    # 全域木で par 以下の辺に向きをつける
    for v in st[par]:
        if st[v]:
            dfs(v)
        if outdeg[v] % 2 == 0:
            ans.append((par, v))
            outdeg[par] += 1
        else:
            ans.append((v, par))
            outdeg[v] += 1


dfs(root)
for r in ans:
    print(*r)