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:
#!/usr/bin/env python3
import sys
from collections import defaultdict, deque
import heapq


def solve(N: int, M: int, A: "List[int]", B: "List[int]"):
    if M % 2 != 0:
        print(-1)
        return

    adj = defaultdict(set)
    for a, b in zip(A, B):
        adj[a].add(b)
        adj[b].add(a)
    parent = [None for _ in range(N)]
    children = [set() for _ in range(N)]

    res = set()
    count = [0 for _ in range(N)]

    Q = deque([0], N)
    visited = [False for _ in range(N)]
    visited[0] = True

    while len(Q) > 0:
        u = Q.popleft()
        for v in adj[u]:
            if parent[u] == v:
                continue
            if visited[v]:
                if (v, u) in res:
                    continue
                res.add((u, v))
                count[u] += 1
                continue
            parent[v] = u
            children[u].add(v)
            visited[v] = True
            Q.append(v)
    visited = [False for _ in range(N)]
    Q = deque([0], N)
    while len(Q) > 0:
        u = Q.popleft()
        if (len(children[u]) > 0
            and not all(visited[v] for v in children[u])
            ):
            Q.appendleft(u)
            Q.extendleft(children[u])
            continue
        
        if parent[u] is None:
            visited[u] = True
            continue
        
        v = parent[u]
        if count[u] % 2 == 0:
            res.add((v, u))
            count[v] += 1
        else:
            res.add((u, v))
            count[u] += 1
        visited[u] = True

    for c, d in res:
        print(c+1, d+1)

    return


# Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    M = int(next(tokens))  # type: int
    A = [int()] * (M)  # type: "List[int]" 
    B = [int()] * (M)  # type: "List[int]" 
    for i in range(M):
        A[i] = int(next(tokens)) - 1
        B[i] = int(next(tokens)) - 1
    solve(N, M, A, B)

if __name__ == '__main__':
    main()