#include <bits/stdc++.h>
using namespace std;
const int maxn=100005;
int n,m,x_[maxn],y_[maxn],ans[maxn],head[maxn],tot=1;
bool vis[maxn],deg[maxn],is_e[maxn];
struct node { int to,next; bool dir; } e[maxn<<1];
inline void addedge(int x,int y,bool dir)
{
	e[++tot].to=y;
	e[tot].dir=dir;
	e[tot].next=head[x];
	head[x]=tot;
}
void dfs1(int x)
{
	vis[x]=1;
	for(int i=head[x];i;i=e[i].next)
		if(!vis[e[i].to])
			is_e[i>>1]=1,dfs1(e[i].to);
}
void dfs2(int x,int ff)
{
	for(int i=head[x];i;i=e[i].next)
		if(is_e[i>>1] && e[i].to!=ff)
		{
			dfs2(e[i].to,x);
			if(deg[e[i].to]) ans[i>>1]=e[i].dir^1,deg[e[i].to]^=1;
			else ans[i>>1]=e[i].dir,deg[x]^=1;
		}
}
int main()
{
	scanf("%d%d",&n,&m);
	if(m&1) return printf("-1"),0;
	for(int i=1;i<=m;i++)
		scanf("%d%d",&x_[i],&y_[i]),addedge(x_[i],y_[i],0),addedge(y_[i],x_[i],1);
	dfs1(1);
	for(int i=1;i<=m;i++)
		if(!is_e[i]) deg[x_[i]]^=1,ans[i]=0;
	dfs2(1,0);
	for(int i=1;i<=m;i++)
		if(!ans[i]) printf("%d %d\n",x_[i],y_[i]);
		else printf("%d %d\n",y_[i],x_[i]);
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 maxn=100005;
int n,m,x_[maxn],y_[maxn],ans[maxn],head[maxn],tot=1;
bool vis[maxn],deg[maxn],is_e[maxn];
struct node { int to,next; bool dir; } e[maxn<<1];
inline void addedge(int x,int y,bool dir)
{
	e[++tot].to=y;
	e[tot].dir=dir;
	e[tot].next=head[x];
	head[x]=tot;
}
void dfs1(int x)
{
	vis[x]=1;
	for(int i=head[x];i;i=e[i].next)
		if(!vis[e[i].to])
			is_e[i>>1]=1,dfs1(e[i].to);
}
void dfs2(int x,int ff)
{
	for(int i=head[x];i;i=e[i].next)
		if(is_e[i>>1] && e[i].to!=ff)
		{
			dfs2(e[i].to,x);
			if(deg[e[i].to]) ans[i>>1]=e[i].dir^1,deg[e[i].to]^=1;
			else ans[i>>1]=e[i].dir,deg[x]^=1;
		}
}
int main()
{
	scanf("%d%d",&n,&m);
	if(m&1) return printf("-1"),0;
	for(int i=1;i<=m;i++)
		scanf("%d%d",&x_[i],&y_[i]),addedge(x_[i],y_[i],0),addedge(y_[i],x_[i],1);
	dfs1(1);
	for(int i=1;i<=m;i++)
		if(!is_e[i]) deg[x_[i]]^=1,ans[i]=0;
	dfs2(1,0);
	for(int i=1;i<=m;i++)
		if(!ans[i]) printf("%d %d\n",x_[i],y_[i]);
		else printf("%d %d\n",y_[i],x_[i]);
	return 0;
}
```