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:
from heapq import heappush, heappop

N, M = map(int, input().split())
A, B = (
    zip(*(map(int, input().split()) for _ in range(M))) if M else
    ((), ())
)


class UnionFindTree:
    def __init__(self, n):
        self.p = [i for i in range(n + 1)]
        self.r = [1 for _ in range(n + 1)]

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, x, y):
        px = self.find(x)
        py = self.find(y)
        if self.r[px] < self.r[py]:
            self.r[py] += self.r[px]
            self.p[px] = py
        else:
            self.r[px] += self.r[py]
            self.p[py] = px

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def size(self, x):
        return self.r[self.find(x)]


# Mが奇数ならば構築不可
# Mが偶数ならば適当に木を構築し、深さの大きい頂点を端点とする辺から向きを決めていく
utf = UnionFindTree(N)
T = [{} for _ in range(N + 1)]
for a, b in zip(A, B):
    if not utf.same(a, b):
        T[a][b] = 1
        T[b][a] = 1
        utf.union(a, b)
INF = 10**9


def dijkstra(G, s):
    dp = [INF for _ in range(len(G))]
    q = []
    heappush(q, (0, s))
    while q:
        c, i = heappop(q)
        if dp[i] == INF:
            dp[i] = c
            for j, w in G[i].items():
                heappush(q, (c + w, j))
    return dp


r = dijkstra(T, 1)
p = [0, 1] + [
    min(T[i], key=lambda j: r[j])
    for i in range(2, N + 1)
]
dp = [0 for _ in range(N + 1)]
for a, b in zip(A, B):
    if b not in T[a]:
        dp[a] += 1
for i in sorted(
        range(1, N + 1),
        key=lambda j: r[j],
        reverse=True
):
    dp[i] += sum(
        (dp[j] + 1) % 2
        for j in T[i]
        if j != p[i]
    )
ans = (
    -1 if M % 2 == 1 else
    '\n'.join(
        '{} {}'.format(a, b)
        if b not in T[a] or (
            dp[a] % 2 == 1 and p[a] == b or
            dp[b] % 2 == 0 and p[b] == a
        ) else
        '{} {}'.format(b, a)
        for a, b in zip(A, B)
    )
)

print(ans)