#include<cstdio>
#include<cstring>
#include<vector>
using std::vector;
const int N=1e5+5;
int n,m;
vector<int> e[N];
int book[N],deg[N];
void dfs1(int x,int pre)
{
	book[x]=book[pre]+1;
	for(int y:e[x])
		if(y==pre)
			continue;
		else if(book[y])
		{
			if(book[y]<book[x])
				deg[y]^=1;
		}
		else
		{
			dfs1(y,x);
			if(deg[y])
				deg[y]^=1;
			else
				deg[x]^=1;
		}
	return;
}
void dfs(int x,int pre)
{
	book[x]=book[pre]+1;
	for(int y:e[x])
		if(y==pre)
			continue;
		else if(book[y])
		{
			if(book[y]<book[x])
				printf("%d %d\n",y,x),deg[y]^=1;
		}
		else
		{
			dfs(y,x);
			if(deg[y])
				printf("%d %d\n",y,x),deg[y]^=1;
			else
				printf("%d %d\n",x,y),deg[x]^=1;
		}
	return;
}
signed main()
{
	int x,y;
	register int i;
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs1(1,0);
	if(deg[1])
		puts("-1");
	else
		memset(book,0,sizeof(book)),dfs(1,0);
	return 0;
}
### 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<cstdio>
#include<cstring>
#include<vector>
using std::vector;
const int N=1e5+5;
int n,m;
vector<int> e[N];
int book[N],deg[N];
void dfs1(int x,int pre)
{
	book[x]=book[pre]+1;
	for(int y:e[x])
		if(y==pre)
			continue;
		else if(book[y])
		{
			if(book[y]<book[x])
				deg[y]^=1;
		}
		else
		{
			dfs1(y,x);
			if(deg[y])
				deg[y]^=1;
			else
				deg[x]^=1;
		}
	return;
}
void dfs(int x,int pre)
{
	book[x]=book[pre]+1;
	for(int y:e[x])
		if(y==pre)
			continue;
		else if(book[y])
		{
			if(book[y]<book[x])
				printf("%d %d\n",y,x),deg[y]^=1;
		}
		else
		{
			dfs(y,x);
			if(deg[y])
				printf("%d %d\n",y,x),deg[y]^=1;
			else
				printf("%d %d\n",x,y),deg[x]^=1;
		}
	return;
}
signed main()
{
	int x,y;
	register int i;
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs1(1,0);
	if(deg[1])
		puts("-1");
	else
		memset(book,0,sizeof(book)),dfs(1,0);
	return 0;
}
```