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;
const int N=1e5+5;
vector<int> e[N];
int vis[N],d[N];
map< pair<int,int>,int > mp;
pair<int,int> mk(int x,int y){
	if(x>y) swap(x,y);
	return make_pair(x,y);
}
void dfs(int x,int fath){
	vis[x]=1; 
//	cout<<x<<endl;
	for(int i=0;i<e[x].size();++i){
		int v=e[x][i];
		if(v==fath) continue;
		if(vis[v]){
			if(mp.count(mk(x,v))) continue;
			mp[mk(x,v)]=1;
			if(d[v]&1)  ++d[v],cout<<v<<" "<<x<<endl;
			else ++d[x],cout<<x<<" "<<v<<endl;
			continue;
		}
		dfs(v,x);
		if(mp.count(mk(x,v))) continue;
		mp[mk(x,v)]=1;
		if(d[v]&1) cout<<v<<" "<<x<<endl,++d[v];
		else cout<<x<<" "<<v<<endl,++d[x];
	}
}
signed main(){
	int n,m,u,v;
	cin>>n>>m;
	if(m&1) return puts("-1"),0;
	for(int i=1;i<=m;++i){
		cin>>u>>v;
		e[u].push_back(v);
		e[v].push_back(u);
	}
	dfs(1,0);
	return 0;
}