#include<bits/stdc++.h>
using namespace std;
const int N=1e6+5;
struct edge{
	int to,next,ok;
} e[N];
int n,m,good=1,h[N];
int visited[N],check[N];
void add(int u,int v){
	e[++good]=(edge){v,h[u],0};
	h[u]=good;
}
void dfs(int u,int p){
	visited[u]=++*visited;
	for(int i=h[u];i;i=e[i].next){
		if(i^p^1){
			if(!visited[e[i].to]){
				dfs(e[i].to,i);
				if(check[e[i].to]&1){
					e[i].ok=-1;
					e[i^1].ok=1;
					check[e[i].to]^=1;
				}
				else{
					e[i].ok=1;
					e[i^1].ok=-1;
					check[u]^=1;
				}
			}
			else if(visited[e[i].to]<visited[u]){
				e[i].ok=1;
				e[i^1].ok=-1;
				check[u]^=1;
			}
		}
	}
}
int main(){
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int u,v;
		cin>>u>>v;
		add(u,v);
		add(v,u);
	}
	for(int i=1;i<=n;i++){
		if(!visited[i]) dfs(i,0);
	}
	for(int i=1;i<=n;i++){
		if(check[i]){
			cout<<"-1";
			return 0;
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=h[i];j;j=e[j].next){
			if(e[j].ok==1){
				cout<<i<<" "<<e[j].to<<endl;
			}
		}
	}
}
### Prompt

Please formulate 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;
const int N=1e6+5;
struct edge{
	int to,next,ok;
} e[N];
int n,m,good=1,h[N];
int visited[N],check[N];
void add(int u,int v){
	e[++good]=(edge){v,h[u],0};
	h[u]=good;
}
void dfs(int u,int p){
	visited[u]=++*visited;
	for(int i=h[u];i;i=e[i].next){
		if(i^p^1){
			if(!visited[e[i].to]){
				dfs(e[i].to,i);
				if(check[e[i].to]&1){
					e[i].ok=-1;
					e[i^1].ok=1;
					check[e[i].to]^=1;
				}
				else{
					e[i].ok=1;
					e[i^1].ok=-1;
					check[u]^=1;
				}
			}
			else if(visited[e[i].to]<visited[u]){
				e[i].ok=1;
				e[i^1].ok=-1;
				check[u]^=1;
			}
		}
	}
}
int main(){
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int u,v;
		cin>>u>>v;
		add(u,v);
		add(v,u);
	}
	for(int i=1;i<=n;i++){
		if(!visited[i]) dfs(i,0);
	}
	for(int i=1;i<=n;i++){
		if(check[i]){
			cout<<"-1";
			return 0;
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=h[i];j;j=e[j].next){
			if(e[j].ok==1){
				cout<<i<<" "<<e[j].to<<endl;
			}
		}
	}
}
```