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
sys.setrecursionlimit(1000000)

N, M = map(int, input().split())
if M % 2 == 1:
  print(-1)
  exit()
  
E = []
A = [[] for _ in range(N)]
for _ in range(M):
  i, j = map(int, input().split())
  E.append([i-1, j-1])
  A[i-1].append(j-1)
  A[j-1].append(i-1)

tree = [[] for _ in range(N)]
visited = [False] * N
queue = [0]
visited[0] = True
while queue:
  i = queue.pop()
  for j in A[i]:
    if not visited[j]:
      tree[i].append(j)
      tree[j].append(i)
      visited[j] = True
      queue.append(j)

D = [0] * N
for i, j in E:
  if i not in tree[j]:
    print(i+1, j+1)
    D[i] += 1

def f(parent, i):
  for j in tree[i]:
    if j == parent:
      continue
    f(i, j)
    if D[j] % 2 == 0:
      print(i+1, j+1)
      D[i] += 1
    else:
      print(j+1, i+1)
      D[j] += 1
  return

f(0, 0)