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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <functional>
#include <string>
#include <cmath>
#include <cassert>
#define SIZE 100005

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

vector <int> vec[SIZE];
bool use[SIZE];
int dep[SIZE];
int deg[SIZE];

void dfs(int v=0,int p=-1,int d=0)
{
	use[v]=true;
	dep[v]=d;
	for(int i=0;i<vec[v].size();i++)
	{
		int to=vec[v][i];
		if(to!=p)
		{
			if(!use[to]) dfs(to,v,d+1);
			else if(dep[to]<dep[v])
			{
				printf("%d %d\n",v+1,to+1);
				deg[v]++;
			}
		}
	}
	if(p!=-1)
	{
		if(deg[v]%2==1)
		{
			printf("%d %d\n",v+1,p+1);
			deg[v]++;
		}
		else
		{
			printf("%d %d\n",p+1,v+1);
			deg[p]++;
		}
	}
}
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	for(int i=0;i<m;i++)
	{
		int a,b;
		scanf("%d %d",&a,&b);a--,b--;
		vec[a].push_back(b);
		vec[b].push_back(a);
	}
	if(m%2==1)
	{
		puts("-1");
		return 0;
	}
	dfs();
	return 0;
}