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
input = sys.stdin.readline
sys.setrecursionlimit(10**6)
from collections import deque

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

# representation matrix
repn = [[] for j in range(n)]
for a, b in AB:
    a -= 1
    b -= 1
    repn[a].append(b)
    repn[b].append(a)


# Minimum Spanning Tree
tree = [set() for i in range(n)]
visited = [False] * n
visited[0] = True
q = deque([0])

while q:
    u = q.popleft()
    for v in repn[u]:
        if visited[v]:
            continue
        visited[v] = True
        tree[u].add(v)
        tree[v].add(u)
        q.append(v)


# define directions of the edges other than that of MST
degree = [0] * n
ans = []

for a, b in AB:
    a -= 1
    b -= 1
    if b in tree[a]:
        continue
    ans.append('{} {}'.format(a+1, b+1))
    degree[a] += 1


# defining orientations
def dfs(u = 0, parent = None):
    for v in tree[u]:
        if v == parent:
            continue
        dfs(v, u)
        if degree[v] % 2 == 0:
            ans.append('{} {}'.format(u+1, v+1))
            degree[u] += 1
        else:
            ans.append('{} {}'.format(v+1, u+1))
            degree[v] += 1

dfs()

if degree[0] % 2 == 1:
    print(-1)
else:
    print('\n'.join(ans))