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 sys import setrecursionlimit
setrecursionlimit(10**8)

N, M = [int(i) for i in input().split()]

if M % 2 == 1:
    print(-1)
    exit()

E = [[] for i in range(N + 1)]
used_e = [False] * M
ans = []

for i in range(M):
    A, B = map(int, input().split())
    E[A].append((B, i))
    E[B].append((A, i))

visited = [False] * (N + 1)


def dfs(prev, v):
    if visited[v]: return
    visited[v] = True
    cnt = 0

    for to, i in E[v]:
        dfs(v, to)
        if not used_e[i]: cnt += 1

    for to, i in E[v]:
        if cnt % 2 == 1 and to == prev: continue
        if not used_e[i]:
            ans.append((v, to))
            used_e[i] = True


dfs(0, 1)

for a in ans: print(*a)