#include <bits/stdc++.h>
using namespace std;
const int N=1e5+100;
int n,m,dg[N],vi[N],tag[N];
int tot,first[N],nxt[N*2],point[N*2],id[N*2];
struct node
{
	int u,v,dir;
}e[N];
inline void add_edge(int x,int y,int z)
{
	tot++;
	nxt[tot]=first[x];
	first[x]=tot;
	point[tot]=y;
	id[tot]=z;
}
void dfs(int x,int last)
{
	vi[x]=1;
	if (dg[x]&1) tag[x]++;
	for (int i=first[x];i!=-1;i=nxt[i])
	{
		int u=point[i];
		if (vi[u]) continue;
		dfs(u,id[i]);
		tag[x]+=tag[u];
	}
	if (tag[x]&1) e[last].dir^=1;
}
int main()
{
	tot=-1;
	memset(first,-1,sizeof(first));
	memset(nxt,-1,sizeof(nxt));
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d",&e[i].u,&e[i].v);
		dg[e[i].u]++;
		e[i].dir=1;
		add_edge(e[i].u,e[i].v,i);
		add_edge(e[i].v,e[i].u,i);
	}
	int cnt=0;
	for (int i=1;i<=n;i++) cnt+=(dg[i]&1);
	if (cnt&1)
	{
		printf("-1\n");
		return 0;
	}
	dfs(1,0);
	for (int i=1;i<=m;i++) if (e[i].dir==0) swap(e[i].u,e[i].v);
	for (int i=1;i<=m;i++) printf("%d %d\n",e[i].u,e[i].v);
}
### 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 N=1e5+100;
int n,m,dg[N],vi[N],tag[N];
int tot,first[N],nxt[N*2],point[N*2],id[N*2];
struct node
{
	int u,v,dir;
}e[N];
inline void add_edge(int x,int y,int z)
{
	tot++;
	nxt[tot]=first[x];
	first[x]=tot;
	point[tot]=y;
	id[tot]=z;
}
void dfs(int x,int last)
{
	vi[x]=1;
	if (dg[x]&1) tag[x]++;
	for (int i=first[x];i!=-1;i=nxt[i])
	{
		int u=point[i];
		if (vi[u]) continue;
		dfs(u,id[i]);
		tag[x]+=tag[u];
	}
	if (tag[x]&1) e[last].dir^=1;
}
int main()
{
	tot=-1;
	memset(first,-1,sizeof(first));
	memset(nxt,-1,sizeof(nxt));
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d",&e[i].u,&e[i].v);
		dg[e[i].u]++;
		e[i].dir=1;
		add_edge(e[i].u,e[i].v,i);
		add_edge(e[i].v,e[i].u,i);
	}
	int cnt=0;
	for (int i=1;i<=n;i++) cnt+=(dg[i]&1);
	if (cnt&1)
	{
		printf("-1\n");
		return 0;
	}
	dfs(1,0);
	for (int i=1;i<=m;i++) if (e[i].dir==0) swap(e[i].u,e[i].v);
	for (int i=1;i<=m;i++) printf("%d %d\n",e[i].u,e[i].v);
}
```