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:
#
# 

n, m = map(int, input().split())
g    = {}
ori  = []

for _ in range(m):
    u, v = map(int, input().split()) 
    ori.append((u, v))
    
    if u not in g:
        g[u] = []
    g[u].append(v)
    
    if v not in g:
        g[v] = []
    g[v].append(u)    
    
def solve(g, ori):
    p    = {}    
    p[1] = 1
    i    = 0
    S    = [1]

    used = [False] * (n+1)
    cnt  = [0]     * (n+1) 
    edge = []
    
    used[1] = True
    
    while i < len(S):
        cur     = S[i]
        
        for u in g[cur]:
            if used[u] == True: continue
            
            used[u] = True
            p[u] = cur
            S.append(u)
        i+=1
    
    for u, v in ori:
        if p[u] != v and p[v] != u:
            cnt[u] += 1
            edge.append((u, v))
        
    for u in S[::-1][:-1]:
        if cnt[u] % 2 == 0:
            edge.append((p[u], u))
            cnt[p[u]] += 1
        else:
            edge.append((u, p[u]))
            
            
    #print(S)
    #print(g)
    #print(p)
    return edge

s = ''
if m % 2 == 1:
    print('-1')
else:
    edge = solve(g, ori)
    for x, y in edge:
        s += str(x)+ ' '+str(y)+'\n'
    print(s)