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:
mod = 1000000007
eps = 10**-9


def main():
    import sys
    from collections import deque
    input = sys.stdin.readline

    N, M = map(int, input().split())
    adj = [[] for _ in range(N+1)]

    if M&1:
        print(-1)
        exit()

    ans = []
    E2i = {}
    num = [0] * (N+1)
    for i in range(M):
        a, b = map(int, input().split())
        adj[a].append(b)
        adj[b].append(a)
        ans.append((a, b))
        E2i[a*(N+1)+b] = i
        E2i[b*(N+1)+a] = i
        num[a] += 1

    que = deque()
    que.append(1)
    seen = [-1] * (N+1)
    seen[1] = 0
    par = [0] * (N+1)
    child = [[] for _ in range(N+1)]
    seq = []
    while que:
        v = que.popleft()
        seq.append(v)
        for u in adj[v]:
            if seen[u] == -1:
                seen[u] = seen[v] + 1
                par[u] = v
                child[v].append(u)
                que.append(u)
    seq.reverse()

    for v in seq:
        if v == 1:
            continue
        if num[v] & 1:
            p = par[v]
            i = E2i[v*(N+1)+p]
            if ans[i] == (p, v):
                ans[i] = (v, p)
                num[v] += 1
                num[p] -= 1
            else:
                ans[i] = (p, v)
                num[p] += 1
                num[v] -= 1

    for c, d in ans:
        print(c, d)


if __name__ == '__main__':
    main()