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**6)

int1 = lambda x: int(x) - 1
n, m = map(int, input().split())
if m % 2:
  print(-1); exit(0)

l = [tuple(map(int1, input().split())) for _ in range(m)]
G = [list() for _ in range(n)]

ps = [-1]*n
tot = [0]*n

def find(x):
  if ps[x] < 0:
    return x
  ps[x] = find(ps[x])
  return ps[x]

def unite(u, v):
  u = find(u)
  v = find(v)
  if ps[u] > ps[v]:
    u, v = v, u
  ps[u] += ps[v]
  ps[v] = u
  return

def same(u, v):
  return find(u) == find(v)

for u, v in l:
  if same(u, v):
    print(u+1, v+1)
    tot[u] += 1
  else:
    G[u].append(v)
    G[v].append(u)
    unite(u, v)

def dfs(v, p):
  for u in G[v]:
    if u == p: continue
    dfs(u, v)
  if p >= 0:
    if tot[v] % 2:
      print(v+1, p+1)
      tot[v] += 1
    else:
      print(p+1, v+1)
      tot[p] += 1
  return

dfs(0, -1)