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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int edge_num=-1;
struct edge
{
	int to,p,nxt;
}e[1000001];
int head[1000001];
void add_edge(int u,int v)
{
	e[++edge_num].to=v;
	e[edge_num].nxt=head[u];
	head[u]=edge_num;
}
int cnt;
int dfn[1000001];
int out[1000001];
void dfs(int u,int fa)
{
	dfn[u]=++cnt;
	for(int i=head[u];i!=-1;i=e[i].nxt)
	{
		int to=e[i].to;
		if(dfn[to]==0)
		{
			dfs(to,u);
			if(out[to]%2==0)
				e[i].p=1,e[i^1].p=-1,out[u]++;
			else
				e[i^1].p=1,e[i].p=-1,out[to]++;
		}
		else if(dfn[to]>dfn[u])
		{
			e[i].p=1;
			e[i^1].p=-1;
			out[u]++;
		}
	}
}
int main()
{
	memset(head,-1,sizeof(head));
	int n,m;
	scanf("%d%d",&n,&m);
	if(m%2==1)
	{
		printf("-1");
		return 0;
	}
	for(int i=1;i<=m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add_edge(u,v);
		add_edge(v,u);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
		for(int j=head[i];j!=-1;j=e[j].nxt)
			if(e[j].p==1)
				printf("%d %d\n",i,e[j].to);
    return 0;
}
/*
5 6
1 2
2 3
3 4
4 5
1 3
1 4
*/