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<sstream>
#include<fstream>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#include<functional>
#include<utility>
#include<bitset>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cstdio>
using namespace std;
vector<int>g[101010];
vector<pair<int,int> >e;
int n,m;
bool vis[101010],d[101010];
void dfs(int x,int y)
{
	if(vis[y])
	{
		return;
	}
	vis[y]=true;
	int z=1;
	if(x>0)
	{
		z=x^e[y].first^e[y].second;
	}
	for(int i=0;i<g[z].size();i++)
	{
		if(g[z][i]!=y)
		{
			dfs(z,g[z][i]);
		}
	}
	if(!x)
	{
		return;
	}
	if(d[z])
	{
		cout<<z<<' '<<x<<endl;
		d[z]=false;
	}
	else
	{
		cout<<x<<' '<<z<<endl;
		d[x]^=true;
	}
}
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;
		e.push_back(make_pair(a,b));
		g[a].push_back(i);
		g[b].push_back(i);
	}
	dfs(0,m);
	return 0;
}