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 maxn=1e5+10;

int n,m,u,v,cnt=0,ans[maxn][2],dep[maxn];
vector<int> E[maxn];
bool Vis[maxn];

void ad(int u,int v){
	ans[cnt][0]=u;
	ans[cnt][1]=v;
	++cnt;
}

bool DFS(int o,int fa){
	Vis[o]=true;
	int lft=0;
	for(int i=0,_=E[o].size();i<_;i++){
		int v=E[o][i];
		if(v==fa)continue;
		if(dep[v]==0){
			dep[v]=dep[o]+1;
			bool hs=DFS(v,o);
			if(hs){
				if(lft!=0){
					ad(o,v);
					ad(o,lft);
					lft=0;
				}else{
					lft=v;
				}
			}
		}else if(dep[v]<dep[o]){
			if(lft!=0){
				ad(o,v);
				ad(o,lft);
				lft=0;
			}else{
				lft=v;
			}
		}
	}
	if(lft!=0){
		ad(o,lft);
		ad(o,fa);
		return false;
	}
	return true;
}

int main(){
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		cin>>u>>v;
		E[u].push_back(v);
		E[v].push_back(u);
	}
	bool flag=true;
	for(int i=1;i<=n&&flag;i++){
		if(!Vis[i]){
			dep[i]=1;
			if(!DFS(i,0))flag=false;
		}
	}
	if(flag){
		for(int i=0;i<cnt;i++){
			cout<<ans[i][0]<<' '<<ans[i][1]<<endl;
		}
	}else{
		cout<<-1;
	}
	return 0;
}