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(10**9)

def dfs1(v):
  for u in g[v]:
    if checked[u]==0:
      checked[u]=1
      g2[v].append(u)
      g3[v].append(u)
      g3[u].append(v)
      dfs1(u)
      
def dfs2(v):
  for u in g2[v]:
    if checked[u]==0:
      checked[u]=1
      dfs2(u)
      if cnt[u]%2==1:
        ans.append((u,v))
        cnt[u]+=1
      else:
        ans.append((v,u))
        cnt[v]+=1
      
n,m=map(int,input().split())
g=[[] for _ in range(n+1)]
g2=[[] for _ in range(n+1)]
g3=[[] for _ in range(n+1)]
cnt=[0]*(n+1)
ans=[]
ansset=set()
for _ in range(m):
  a,b=map(int,input().split())
  g[a].append(b)
  g[b].append(a)
if m%2==1:
  print(-1)
else:
  checked=[0]*(n+1)
  checked[1]=1
  dfs1(1)
  for v in range(1,n+1):
    for u in g[v]:
      if u not in g3[v]:
        if (u,v) not in ansset:
          ans.append((v,u))
          ansset.add((v,u))
          cnt[v]+=1
  checked=[0]*(n+1)
  checked[1]=1
  dfs2(1)
  for v,u in ans:
    print(v,u)