#include<bits/stdc++.h>
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#define for1(i,l,r) for(int i=(l);i<(r);i++)
#define for2(i,l,r) for(int i=(l);i>(r);i--)
using namespace std;
int n,m,u,v,root,le;
vector<vector<pair<int,int> > > r;
vector<pair<int,int> > ans;
int cnt[100006],vis[100006],out[100006];
void dfs(int u){
	vis[u]=true;
	cnt[u]+=out[u];
	for(int i=0;i<r[u].size();i++){
		int x=r[u][i].first,y=r[u][i].second;
		if(!vis[x]){
			dfs(x);
			cnt[u]+=cnt[x];
			cnt[u]%=2;
			if(cnt[x]%2==1) swap(ans[y].first,ans[y].second);
		}
	}
}
signed main(){
//	freopen(".inp","r",stdin);
//	freopen(".out","w",stdout);
//	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n>>m;
	r.assign(n+1,vector<pair<int,int> >());
	for(int i=0;i<m;i++){
		cin>>u>>v;
		out[u]++;
		r[u].push_back({v,i});
		r[v].push_back({u,i});
		ans.push_back({u,v});
	}
	for(int i=1;i<=n;i++){
		out[i]%=2;
		if(out[i]) le++;
	}
	if(le%2==1)	cout<<-1;
	else{
		dfs(1);
		for(int i=0;i<m;i++) cout<<ans[i].first<<" "<<ans[i].second<<endl;
	}
}

### 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>
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#define for1(i,l,r) for(int i=(l);i<(r);i++)
#define for2(i,l,r) for(int i=(l);i>(r);i--)
using namespace std;
int n,m,u,v,root,le;
vector<vector<pair<int,int> > > r;
vector<pair<int,int> > ans;
int cnt[100006],vis[100006],out[100006];
void dfs(int u){
	vis[u]=true;
	cnt[u]+=out[u];
	for(int i=0;i<r[u].size();i++){
		int x=r[u][i].first,y=r[u][i].second;
		if(!vis[x]){
			dfs(x);
			cnt[u]+=cnt[x];
			cnt[u]%=2;
			if(cnt[x]%2==1) swap(ans[y].first,ans[y].second);
		}
	}
}
signed main(){
//	freopen(".inp","r",stdin);
//	freopen(".out","w",stdout);
//	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n>>m;
	r.assign(n+1,vector<pair<int,int> >());
	for(int i=0;i<m;i++){
		cin>>u>>v;
		out[u]++;
		r[u].push_back({v,i});
		r[v].push_back({u,i});
		ans.push_back({u,v});
	}
	for(int i=1;i<=n;i++){
		out[i]%=2;
		if(out[i]) le++;
	}
	if(le%2==1)	cout<<-1;
	else{
		dfs(1);
		for(int i=0;i<m;i++) cout<<ans[i].first<<" "<<ans[i].second<<endl;
	}
}

```