#include<bits/stdc++.h>
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define pii pair<int,int>
#define fi first
#define se second
#define PB push_back
using namespace std;
const int N=100005;
struct edge{
	int to,next,ok;
}e[N*2];
int head[N],tot=1,n,m;
int vis[N],deg[N];
void add(int x,int y){
	e[++tot]=(edge){y,head[x],0};
	head[x]=tot;
}
void dfs(int x,int fae){
	vis[x]=++*vis;
	for (int i=head[x];i;i=e[i].next)
		if ((i^fae^1))
			if (!vis[e[i].to]){
				dfs(e[i].to,i);
				if (deg[e[i].to]&1){	
					e[i].ok=-1;
					e[i^1].ok=1;
					deg[e[i].to]^=1;
				}
				else{
					e[i].ok=1;
					e[i^1].ok=-1;
					deg[x]^=1;
				}
			}
			else if (vis[e[i].to]<vis[x]){
				e[i].ok=1;
				e[i^1].ok=-1;
				deg[x]^=1;
			}
}
int main(){
	scanf("%d%d",&n,&m);
	For(i,1,m){
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	For(i,1,n)
		if (!vis[i])
			dfs(i,0);
	For(i,1,n)
		if (deg[i])
			return puts("-1"),0;
	For(i,1,n)
		for (int j=head[i];j;j=e[j].next)
			if (e[j].ok==1)
				printf("%d %d\n",i,e[j].to);
}
### 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<bits/stdc++.h>
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define pii pair<int,int>
#define fi first
#define se second
#define PB push_back
using namespace std;
const int N=100005;
struct edge{
	int to,next,ok;
}e[N*2];
int head[N],tot=1,n,m;
int vis[N],deg[N];
void add(int x,int y){
	e[++tot]=(edge){y,head[x],0};
	head[x]=tot;
}
void dfs(int x,int fae){
	vis[x]=++*vis;
	for (int i=head[x];i;i=e[i].next)
		if ((i^fae^1))
			if (!vis[e[i].to]){
				dfs(e[i].to,i);
				if (deg[e[i].to]&1){	
					e[i].ok=-1;
					e[i^1].ok=1;
					deg[e[i].to]^=1;
				}
				else{
					e[i].ok=1;
					e[i^1].ok=-1;
					deg[x]^=1;
				}
			}
			else if (vis[e[i].to]<vis[x]){
				e[i].ok=1;
				e[i^1].ok=-1;
				deg[x]^=1;
			}
}
int main(){
	scanf("%d%d",&n,&m);
	For(i,1,m){
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	For(i,1,n)
		if (!vis[i])
			dfs(i,0);
	For(i,1,n)
		if (deg[i])
			return puts("-1"),0;
	For(i,1,n)
		for (int j=head[i];j;j=e[j].next)
			if (e[j].ok==1)
				printf("%d %d\n",i,e[j].to);
}
```