#include<bits/stdc++.h>
using namespace std;

int cnt[100010];

bool vis[100010],vise[100010];
set<int> st;
vector<pair<int,int> > e[100010];
int u[100010],v[100010];
void dfs(int x,int p){
	vis[x]=true;
	for(auto it:e[x]){
		int y = it.first;
		if(!vis[y]){
			vise[it.second]=true;
			dfs(y,x);
		}
	}
}

void dfs2(int x,int p){
	vis[x]=true;
	for(auto it:e[x]){
		int y = it.first;
		if(!vis[y]){
			dfs2(y,x);
		}
	}
	if(p!=0){
		if(cnt[x]){
			cout<<x<<' '<<p<<'\n';
		}
		else{
			cout<<p<<' '<<x<<'\n';
			cnt[p]^=1;
		}
	}
}
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		cin>>u[i]>>v[i];
		e[u[i]].emplace_back(v[i],i);
		e[v[i]].emplace_back(u[i],i);
	}
	if(m&1){
		cout<<-1<<endl;
		return 0;
	}
	dfs(1,0);
	for(int i=1;i<=m;i++)
		if(!vise[i]){
			cout<<u[i]<<' '<<v[i]<<'\n';
			cnt[u[i]]^=1;
		}
	memset(vis,0,sizeof vis);
	dfs2(1,0);
}

### Prompt

Your challenge is to write a CPP solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

int cnt[100010];

bool vis[100010],vise[100010];
set<int> st;
vector<pair<int,int> > e[100010];
int u[100010],v[100010];
void dfs(int x,int p){
	vis[x]=true;
	for(auto it:e[x]){
		int y = it.first;
		if(!vis[y]){
			vise[it.second]=true;
			dfs(y,x);
		}
	}
}

void dfs2(int x,int p){
	vis[x]=true;
	for(auto it:e[x]){
		int y = it.first;
		if(!vis[y]){
			dfs2(y,x);
		}
	}
	if(p!=0){
		if(cnt[x]){
			cout<<x<<' '<<p<<'\n';
		}
		else{
			cout<<p<<' '<<x<<'\n';
			cnt[p]^=1;
		}
	}
}
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		cin>>u[i]>>v[i];
		e[u[i]].emplace_back(v[i],i);
		e[v[i]].emplace_back(u[i],i);
	}
	if(m&1){
		cout<<-1<<endl;
		return 0;
	}
	dfs(1,0);
	for(int i=1;i<=m;i++)
		if(!vise[i]){
			cout<<u[i]<<' '<<v[i]<<'\n';
			cnt[u[i]]^=1;
		}
	memset(vis,0,sizeof vis);
	dfs2(1,0);
}

```