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;
#define N 100005
int n,m,fa[N],out[N];
vector<int>g[N];
inline int Find(int x)
{
	return x==fa[x]?x:fa[x]=Find(fa[x]);
}
inline void dfs(int x,int fa)
{
	for(int i=0;i<g[x].size();i++)
	{
		int v=g[x][i];
		if(v==fa)
			continue;
		dfs(v,x);
		if(out[v]&1)
			printf("%d %d\n",v,x);
		else
		{
			out[x]++;
			printf("%d %d\n",x,v);
		}	
	}
	return;
}
int main()
{
	scanf("%d%d",&n,&m);
	if(m&1)
	{
		printf("-1");
		return 0;
	}
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		if(Find(u)!=Find(v))
		{
			g[u].push_back(v);
			g[v].push_back(u);
			fa[Find(u)]=Find(v);
		}
		else
		{
			out[v]++;
			printf("%d %d\n",v,u);
		}
	}
	dfs(1,0);
	return 0;
}