#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=100100;
vector<int> es[N];
int d[N]={0};
void ae(int u,int v){
	es[u].push_back(v);
}
struct UFS{
	int fa[N];
	void init(int n){
		for(int i=0;i<=n;i++)fa[i]=i;
	}
	int find(int x){
		return fa[x]==x?x:fa[x]=find(fa[x]);
	}
	bool merge(int a,int b){
		a=find(a);
		b=find(b);
		if(a==b)return false;
		fa[a]=b;
		return true;
	}
}ufs;
void dfs(int x,int fa=-1){
	for(vector<int>::iterator it=es[x].begin();it!=es[x].end();++it){
		if(*it!=fa){
			dfs(*it,x);
		}
	}
	if(fa!=-1){
		if(d[x]==1){
			printf("%d %d\n",x,fa);
			d[x]^=1;
		}
		else{
			printf("%d %d\n",fa,x);
			d[fa]^=1;
		}
	}
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	if(m%2!=0){
		printf("-1\n");
		exit(0);
	}
	ufs.init(n);
	while(m--){
		int u,v;
		scanf("%d%d",&u,&v);
		if(!ufs.merge(u,v)){
			printf("%d %d\n",u,v);
			d[u]^=1;
		}
		else{
			ae(u,v);
			ae(v,u);
		}
	}
	dfs(1);
	return 0;
}
### Prompt

Your task is to create 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;
typedef long long LL;
const int N=100100;
vector<int> es[N];
int d[N]={0};
void ae(int u,int v){
	es[u].push_back(v);
}
struct UFS{
	int fa[N];
	void init(int n){
		for(int i=0;i<=n;i++)fa[i]=i;
	}
	int find(int x){
		return fa[x]==x?x:fa[x]=find(fa[x]);
	}
	bool merge(int a,int b){
		a=find(a);
		b=find(b);
		if(a==b)return false;
		fa[a]=b;
		return true;
	}
}ufs;
void dfs(int x,int fa=-1){
	for(vector<int>::iterator it=es[x].begin();it!=es[x].end();++it){
		if(*it!=fa){
			dfs(*it,x);
		}
	}
	if(fa!=-1){
		if(d[x]==1){
			printf("%d %d\n",x,fa);
			d[x]^=1;
		}
		else{
			printf("%d %d\n",fa,x);
			d[fa]^=1;
		}
	}
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	if(m%2!=0){
		printf("-1\n");
		exit(0);
	}
	ufs.init(n);
	while(m--){
		int u,v;
		scanf("%d%d",&u,&v);
		if(!ufs.merge(u,v)){
			printf("%d %d\n",u,v);
			d[u]^=1;
		}
		else{
			ae(u,v);
			ae(v,u);
		}
	}
	dfs(1);
	return 0;
}
```