//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
const int N=5e5+5;
struct eg{int v,nx;}e[N<<1];int head[N],tot=1;
inline void add(R int u,R int v){e[++tot]={v,head[u]},head[u]=tot;}
int n,m,vis[N],ok[N],deg[N];
void dfs(int u,int fa){
	vis[u]=1;
	go(u)if(v!=fa){
		if(!vis[v])dfs(v,u);
		else if(!ok[i])printf("%d %d\n",u,v),ok[i]=ok[i^1]=1,deg[u]^=1;
	}
	if(fa){
		if(deg[u])printf("%d %d\n",u,fa);
		else printf("%d %d\n",fa,u),deg[fa]^=1;
	}
}
int main(){
	scanf("%d%d",&n,&m);
	if(m&1)return puts("-1"),0;
	for(R int i=1,u,v;i<=m;++i)scanf("%d%d",&u,&v),add(u,v),add(v,u);
	dfs(1,0);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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
//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
const int N=5e5+5;
struct eg{int v,nx;}e[N<<1];int head[N],tot=1;
inline void add(R int u,R int v){e[++tot]={v,head[u]},head[u]=tot;}
int n,m,vis[N],ok[N],deg[N];
void dfs(int u,int fa){
	vis[u]=1;
	go(u)if(v!=fa){
		if(!vis[v])dfs(v,u);
		else if(!ok[i])printf("%d %d\n",u,v),ok[i]=ok[i^1]=1,deg[u]^=1;
	}
	if(fa){
		if(deg[u])printf("%d %d\n",u,fa);
		else printf("%d %d\n",fa,u),deg[fa]^=1;
	}
}
int main(){
	scanf("%d%d",&n,&m);
	if(m&1)return puts("-1"),0;
	for(R int i=1,u,v;i<=m;++i)scanf("%d%d",&u,&v),add(u,v),add(v,u);
	dfs(1,0);
	return 0;
}
```