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;
int n,m,par[100005];
vector<int>g[100005];
vector<pair<int,int> >ans;
int deg[100005];
int find(int x){return x==par[x]?x:par[x]=find(par[x]);}
void dfs(int x,int par){
	for(auto to:g[x]){
		if(to!=par){
			dfs(to,x);
		}
	}
	if(deg[x]){
		ans.push_back(make_pair(x,par));
		deg[x]^=1;
	}else{
		ans.push_back(make_pair(par,x));
		deg[par]^=1;
	}
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>m;
	if(m&1){
		cout<<-1<<endl;
		return 0;
	}
	for(int i=1;i<=n;i++)par[i]=i;
	for(int i=1;i<=m;i++){
		int x,y;
		cin>>x>>y;
		if(find(x)!=find(y)){
			par[find(x)]=find(y);
			g[x].push_back(y);
			g[y].push_back(x);
		}else{
			ans.push_back(make_pair(x,y));
			deg[x]^=1;
		}
	}
	dfs(1,0);
	ans.pop_back();
	for(auto p:ans){
		cout<<p.first<<" "<<p.second<<endl;
	}
}