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, deque
from heapq import heappush, heappop
sys.setrecursionlimit(10**6)

def main():
    input = sys.stdin.readline
    N, M = map(int, input().split())
    if M % 2 == 1:
        print(-1)
        return 0

    G = [[] for _ in range(N+1)]
    for _ in range(M):
        a, b = map(int, input().split())
        G[a].append(b)
        G[b].append(a)

    heap = []
    dist = [N] * (N+1)
    dist[1] = 0
    d = 0
    q = deque()
    q.append(1)
    while len(q) > 0:
        d += 1
        q_ = deque()
        while len(q) > 0:
            a = q.popleft()
            for b in G[a]:
                if dist[b] > d:
                    dist[b] = d
                    q_.append(b)
                    heappush(heap, (-d, b))
        q = q_

    cnt = [0] * (N+1)
    visited = [False] * (N+1)
    ans = []
    while len(heap) > 0:
        _, a = heappop(heap)
        visited[a] = True
        for b in G[a]:
            if visited[b]:
                continue

            if cnt[a] % 2 == 1:
                ans.append((a, b))
                cnt[a] += 1
            else:
                ans.append((b, a))
                cnt[b] += 1

    for a, b in ans:
        print('{} {}'.format(a, b))
    exit()


if __name__ == '__main__':
    main()