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;
typedef long long llo;
#define mp make_pair
#define pb push_back
#define a first 
#define b second
//#define endl '\n' 

vector<pair<int,int>> ans;
int n,m;
int val[100001];
int par[100001];
int find(int no){
	if(par[no]==no){
		return no;
	}
	par[no]=find(par[no]);
	return par[no];
}
vector<int> adj[100001];
void dfs(int no,int par=-1){

	for(auto j:adj[no]){
		if(j!=par){
			dfs(j,no);
			if(val[j]==0){
				ans.pb({no,j});
			}
			else{
				ans.pb({j,no});
			}
			val[ans.back().a]^=1;
		}
	}
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin>>n>>m;

	for(int i=0;i<n;i++){
		par[i]=i;
	}
	for(int i=0;i<m;i++){
		int aa,bb;
		cin>>aa>>bb;
		aa--;
		bb--;
		int x=find(aa);
		int y=find(bb);
		if(x!=y){
			par[x]=y;
			adj[aa].pb(bb);
			adj[bb].pb(aa);
		}
		else{
			val[aa]^=1;
			ans.pb({aa,bb});
		}
	}
	if(m%2==1){
		cout<<-1<<endl;
		return 0;
	}
	dfs(0);
	for(auto j:ans){
		cout<<j.a+1<<" "<<j.b+1<<endl;
	}



	return 0;
}