#include<bits/stdc++.h>
using namespace std;

inline int read()
{
	int ret=0; char c=getchar();
	while(c<48||c>57)c=getchar();
	while(c>=48 && c<=57)ret=ret*10+c-48,c=getchar();
	return ret;
}

#define N 200901
int u,v,m,a[N],dfn[N],vi[N],si[N],tt,n,c[N<<1][2],tot,f[N<<1];

void dfs(int p, int pr)
{
	dfn[p]=++tt;
	vi[p]=1;
	for(int o=a[p]; o; o=c[o][1]){
		if(c[o][0]==pr)continue;
		int u=c[o][0];
		if(vi[u]){
			if(dfn[u]<dfn[p]){
				++si[p], f[o]=1;
			}
		} else {
			dfs(u,p);
			if(si[u]%2)f[o^1]=1;
			else ++si[p],f[o]=1;
		}
	}
}

int main()
{
	n=read(); m=read();
	tot=1;
	for(int i=1;i<=m;++i){
		u=read(), v=read();
		c[++tot][0]=v; c[tot][1]=a[u]; a[u]=tot;
		c[++tot][0]=u; c[tot][1]=a[v]; a[v]=tot;
	}
	dfs(1,0);
	if(si[1]%2){
		puts("-1");
		exit(0);
	}
	for(int i=1;i<=m;++i){
		if(f[i<<1])printf("%d %d\n",c[(i<<1)|1][0],c[i<<1][0]);
		else printf("%d %d\n",c[i<<1][0],c[(i<<1)|1][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<bits/stdc++.h>
using namespace std;

inline int read()
{
	int ret=0; char c=getchar();
	while(c<48||c>57)c=getchar();
	while(c>=48 && c<=57)ret=ret*10+c-48,c=getchar();
	return ret;
}

#define N 200901
int u,v,m,a[N],dfn[N],vi[N],si[N],tt,n,c[N<<1][2],tot,f[N<<1];

void dfs(int p, int pr)
{
	dfn[p]=++tt;
	vi[p]=1;
	for(int o=a[p]; o; o=c[o][1]){
		if(c[o][0]==pr)continue;
		int u=c[o][0];
		if(vi[u]){
			if(dfn[u]<dfn[p]){
				++si[p], f[o]=1;
			}
		} else {
			dfs(u,p);
			if(si[u]%2)f[o^1]=1;
			else ++si[p],f[o]=1;
		}
	}
}

int main()
{
	n=read(); m=read();
	tot=1;
	for(int i=1;i<=m;++i){
		u=read(), v=read();
		c[++tot][0]=v; c[tot][1]=a[u]; a[u]=tot;
		c[++tot][0]=u; c[tot][1]=a[v]; a[v]=tot;
	}
	dfs(1,0);
	if(si[1]%2){
		puts("-1");
		exit(0);
	}
	for(int i=1;i<=m;++i){
		if(f[i<<1])printf("%d %d\n",c[(i<<1)|1][0],c[i<<1][0]);
		else printf("%d %d\n",c[i<<1][0],c[(i<<1)|1][0]);
	}
}
```