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
sys.setrecursionlimit(10**6)
N,M = map(int,input().split())
AB = [tuple(map(int,input().split())) for i in range(M)]
if M%2:
    print(-1)
    exit()

class UnionFind:
    def __init__(self,N):
        self.parent = [i for i in range(N)]
        self.rank = [0] * N
        self.count = 0
    def root(self,a):
        if self.parent[a] == a:
            return a
        else:
            self.parent[a] = self.root(self.parent[a])
            return self.parent[a]
    def is_same(self,a,b):
        return self.root(a) == self.root(b)
    def unite(self,a,b):
        ra = self.root(a)
        rb = self.root(b)
        if ra == rb: return
        if self.rank[ra] < self.rank[rb]:
            self.parent[ra] = rb
        else:
            self.parent[rb] = ra
            if self.rank[ra] == self.rank[rb]: self.rank[ra] += 1
        self.count += 1
uf = UnionFind(N)
outdeg = [0] * N
es = [[] for _ in range(N)]
ans = []
for a,b in AB:
    a,b = a-1,b-1
    if uf.is_same(a,b):
        outdeg[a] += 1
        ans.append((a+1,b+1))
    else:
        uf.unite(a,b)
        es[a].append(b)
        es[b].append(a)

def rec(v,p=-1):
    k = outdeg[v]
    for to in es[v]:
        if to==p: continue
        r = rec(to,v)
        if r:
            ans.append((to+1,v+1))
        else:
            ans.append((v+1,to+1))
        k += (r+1)
    return k%2
rec(0)

for row in ans:
    print(*row)