#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
const int N=1e5+5;
int n,m,u[N],v[N],check[N],deg[N];
vector<pair<int,int> > ans;
vector<int> adj[N];
map<pair<int,int>,int> dir;
void dfs(int u,int p){
	check[u]=1;
	for(int v: adj[u]){
		if(check[v]==0){
			dfs(v,u);
		}
	}
	if(deg[u]%2==1){
		if(dir[{u,p}]==1){
			dir[{u,p}]=0;
			dir[{p,u}]=1;
			deg[u]--;
			deg[p]++;
		}else{
			dir[{u,p}]=1;
			dir[{p,u}]=0;
			deg[u]++;
			deg[p]--;
		}
	}
}
int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		cin>>u[i]>>v[i];
		adj[u[i]].push_back(v[i]);
		adj[v[i]].push_back(u[i]);
		dir[{u[i],v[i]}]=1;
		deg[u[i]]++;
	}
	if(m%2==1){
		cout<<"-1";
		return 0;
	}
	dfs(1,1);
	for(int i=1;i<=m;i++){
		if(dir[{u[i],v[i]}]==1){
			cout<<u[i]<<" "<<v[i]<<endl;
		}else{
			cout<<v[i]<<" "<<u[i]<<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>
using namespace std;
#define fi first
#define se second
const int N=1e5+5;
int n,m,u[N],v[N],check[N],deg[N];
vector<pair<int,int> > ans;
vector<int> adj[N];
map<pair<int,int>,int> dir;
void dfs(int u,int p){
	check[u]=1;
	for(int v: adj[u]){
		if(check[v]==0){
			dfs(v,u);
		}
	}
	if(deg[u]%2==1){
		if(dir[{u,p}]==1){
			dir[{u,p}]=0;
			dir[{p,u}]=1;
			deg[u]--;
			deg[p]++;
		}else{
			dir[{u,p}]=1;
			dir[{p,u}]=0;
			deg[u]++;
			deg[p]--;
		}
	}
}
int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		cin>>u[i]>>v[i];
		adj[u[i]].push_back(v[i]);
		adj[v[i]].push_back(u[i]);
		dir[{u[i],v[i]}]=1;
		deg[u[i]]++;
	}
	if(m%2==1){
		cout<<"-1";
		return 0;
	}
	dfs(1,1);
	for(int i=1;i<=m;i++){
		if(dir[{u[i],v[i]}]==1){
			cout<<u[i]<<" "<<v[i]<<endl;
		}else{
			cout<<v[i]<<" "<<u[i]<<endl;
		}
	}
}
```