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**9)

N, M = map(int, input().split())
adjL = [[] for _ in range(N)]
for _ in range(M):
    A, B = map(int, input().split())
    adjL[A-1].append(B-1)
    adjL[B-1].append(A-1)

if M % 2:
    print(-1)
    sys.exit()

def makeSpanningTree(vNow, vPar):
    useds[vNow] = True
    for v2 in adjL[vNow]:
        if v2 == vPar:
            pass
        elif useds[v2]:
            if vNow < v2:
                anss.append((vNow, v2))
                outdegs[vNow] += 1
        else:
            adjL2[vNow].append(v2)
            adjL2[v2].append(vNow)
            makeSpanningTree(v2, vNow)

anss = []
adjL2 = [[] for _ in range(N)]
outdegs = [0] * N
useds = [False] * (N)
makeSpanningTree(0, -1)

def dfs(vNow, vPar):
    for v2 in adjL2[vNow]:
        if v2 != vPar:
            dfs(v2, vNow)
    if outdegs[vNow] % 2:  # 出す
        anss.append((vNow, vPar))
        outdegs[vNow] += 1
    else:  # 入れる
        anss.append((vPar, vNow))
        outdegs[vPar] += 1

dfs(0, -1)
anss.pop()

print('\n'.join(map(lambda x: str(x[0]+1)+' '+str(x[1]+1), anss)))