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<iostream>
#include<vector>
#include<queue>
using namespace std;
struct edge{
	int to,rev,F;
};
vector<edge>G[1<<17];
int N,M;
int vis[1<<17];
void dfs(int u,int p)
{
	vis[u]=1;
	int cnt=0,par=-1;
	for(int i=0;i<G[u].size();i++)
	{
		if(vis[G[u][i].to]==0)dfs(G[u][i].to,u);
		if(G[u][i].to!=p&&G[u][i].F!=0)
		{
			if(G[u][i].F==-1)
			{
				G[u][i].F=1;
				G[G[u][i].to][G[u][i].rev].F=0;
			}
			cnt++;
		}
		else if(G[u][i].to==p)par=i;
	}
	if(cnt%2==1)
	{
		G[u][par].F=1;
		G[G[u][par].to][G[u][par].rev].F=0;
	}
	return;
}
int main()
{
	cin>>N>>M;
	if(M%2==1)
	{
		cout<<-1<<endl;
		return 0;
	}
	for(int i=0;i<M;i++)
	{
		int a,b;cin>>a>>b;
		a--,b--;
		G[a].push_back((edge){b,(int)G[b].size(),-1});
		G[b].push_back((edge){a,(int)G[a].size()-1,-1});
	}
	dfs(0,-1);
	for(int i=0;i<N;i++)
	{
		for(edge&e:G[i])
		{
			if(e.F==1)cout<<i+1<<" "<<e.to+1<<endl;
		}
	}
}