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:
from collections import defaultdict
from collections import deque
from sys import setrecursionlimit
setrecursionlimit(10**9)
N, M = map(int, input().split())
if M % 2 == 1:
  print(-1)
  exit(0)

tree = defaultdict(list)
for i in range(M):
  temp = list(map(int, input().split()))
  tree[temp[0]].append([temp[1], i])
  tree[temp[1]].append([temp[0], i])
usede = [0] * M
visitedv = [0] * (N + 1)
res = []
def dfs(prev, v):
  global usede
  global visitedv
  global res
  
  if visitedv[v] == 1:
    return
  visitedv[v] = 1
  movable = 0
  
  for l in tree[v]:
    w = l[0]
    i = l[1]
    dfs(v, w)
    if usede[i] == 0:
      movable += 1
 
  for l in tree[v]:
    w = l[0]
    i = l[1]
    if w == prev and movable % 2 == 1:
      continue
    if usede[i] == 0:
      res.append((v, w))
      usede[i] = 1
dfs(0, 1)
for v in res:
  print(v[0], v[1])