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
from collections import defaultdict
sys.setrecursionlimit(10**7)
readline = sys.stdin.readline
def solve(p, v, rs):
    als = G[v]
    cs = g[v]
    c = 0
    for u in als:
        if u in cs:
            c += solve(v, u, rs)
        elif u != p:
            if v < u:
                c += 1
            rs.add((min(u,v), max(u,v)))
    if p == None:
        return
    elif c%2 != 0:
        rs.add((v, p))
        return 0
    else:
        rs.add((p,v))
        return 1

N, M = map(int, readline().split())
if M %2 != 0:
    print(-1)
else:
    G = defaultdict(set)
    for v in range(M):
        a, b = map(int, readline().split())
        G[a].add(b)
        G[b].add(a)

    stack=[1]
    vs = set([1])
    g = defaultdict(set)
    counter = 1
    while True:
        v = stack.pop()
        for u in G[v]:
            if u in vs:
                continue
            vs.add(u)
            g[v].add(u)
            stack.append(u)
            counter += 1
        if counter == N:
            break
    rs = set()
    solve(None,1, rs)
    for r in rs:
        print(r[0], r[1])