#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

const int N=100005;

int n,m,cnt,last[N],dep[N],deg[N];
bool vis[N];
struct edge{int to,next;}e[N*2];

void addedge(int u,int v)
{
	e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;
	e[++cnt].to=u;e[cnt].next=last[v];last[v]=cnt;
}

void dfs1(int x,int fa)
{
	vis[x]=1;
	dep[x]=dep[fa]+1;
	for (int i=last[x];i;i=e[i].next)
		if (!vis[e[i].to]) dfs1(e[i].to,x);
		else if (e[i].to!=fa&&dep[e[i].to]>dep[x]) deg[x]^=1,printf("%d %d\n",x,e[i].to);
}

void dfs2(int x,int fa)
{
	vis[x]=1;
	for (int i=last[x];i;i=e[i].next)
		if (!vis[e[i].to]) dfs2(e[i].to,x);
	if (!fa) return;
	if (deg[x]) deg[x]^=1,printf("%d %d\n",x,fa);
	else deg[fa]^=1,printf("%d %d\n",fa,x);
}

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		addedge(x,y);
	}
	if (m&1) {puts("-1");return 0;}
	dfs1(1,0);
	memset(vis,0,sizeof(vis));
	dfs2(1,0);
	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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

const int N=100005;

int n,m,cnt,last[N],dep[N],deg[N];
bool vis[N];
struct edge{int to,next;}e[N*2];

void addedge(int u,int v)
{
	e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;
	e[++cnt].to=u;e[cnt].next=last[v];last[v]=cnt;
}

void dfs1(int x,int fa)
{
	vis[x]=1;
	dep[x]=dep[fa]+1;
	for (int i=last[x];i;i=e[i].next)
		if (!vis[e[i].to]) dfs1(e[i].to,x);
		else if (e[i].to!=fa&&dep[e[i].to]>dep[x]) deg[x]^=1,printf("%d %d\n",x,e[i].to);
}

void dfs2(int x,int fa)
{
	vis[x]=1;
	for (int i=last[x];i;i=e[i].next)
		if (!vis[e[i].to]) dfs2(e[i].to,x);
	if (!fa) return;
	if (deg[x]) deg[x]^=1,printf("%d %d\n",x,fa);
	else deg[fa]^=1,printf("%d %d\n",fa,x);
}

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		addedge(x,y);
	}
	if (m&1) {puts("-1");return 0;}
	dfs1(1,0);
	memset(vis,0,sizeof(vis));
	dfs2(1,0);
	return 0;
}
```