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(100001)

n, m = map(int, input().split())
edges = [[] for _ in range(n)]
visited = [False] * n
for _ in range(m):
    a, b = map(int, input().split())
    la = len(edges[a - 1])
    lb = len(edges[b - 1])
    edges[a - 1].append([b - 1, lb])
    edges[b - 1].append([a - 1, la])

def dfs(p, v):
    visited[v] = True
    for (i, r) in edges[v]:
        if not visited[i] and r != -1 and i != p:
            dfs(v, i)
    odd = False
    pr = -1
    for (i, r) in edges[v]:
        if i != p:
            if r != -1:
                edges[i][r][1] = -1
                odd = not odd
        else:
            pr = r
    if odd:
        edges[p][pr][1] = -1

if m % 2 == 1:
    print(-1)
else:
    dfs(-1, 0)
    for v in range(n):
        for (i, r) in edges[v]:
            if r >= 0:
                print(v + 1, i + 1)