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 deque

n, m = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(m)]

if m % 2:
    print(-1)
    exit()

adj = [[] for _ in range(n + 1)]
for a, b in ab:
    adj[a].append(b)
    adj[b].append(a)

s = 1
dq = deque([s])
d = [-1] * (n + 1)
d[s] = 0
p = [-1] * (n + 1)
order = []
while dq:
    u = dq.popleft()
    order.append(u)
    for v in adj[u]:
        if d[v] == -1:
            d[v] = d[u] + 1
            p[v] = u
            dq.append(v)

out = [0] * (n + 1)
ans = []
for a, b in ab:
    if p[a] == b or p[b] == a:
        continue
    ans.append((a, b))
    out[a] ^= 1

for v in order[1:][::-1]:
    u = p[v]
    if out[v]:
        ans.append((v, u))
        out[v] ^= 1
    else:
        ans.append((u, v))
        out[u] ^= 1

for u, v in ans:
    print(u, v)