#include<bits/stdc++.h>
using namespace std;
const int N=100005;
int T,n,m,i,u[N],v[N];
vector<int>e[N];
int dfn[N],xb,id[N],dad[N],de[N],ans[N];
bool vi[N],ind[N];
void dfs(int x,int fa){
	vi[x]=1;dfn[id[x]=++xb]=x;
	for(int i=0;i<e[x].size();++i){
		int j=e[x][i],y=u[j]^v[j]^x;
		if(ans[j]==-1)ans[j]=x==v[j];
		if(vi[y] && id[y]<id[x] && y!=fa)ind[y]^=1;
		if(!vi[y])dad[y]=x,de[y]=j,dfs(y,x);
	}
}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);
	cin>>n>>m;xb=0;for(i=1;i<=n;++i)e[i].clear();
	memset(ind+1,0,n);memset(ans+1,-1,m<<2);memset(id+1,0,n<<2);memset(vi+1,0,n);
	for(i=1;i<=m;++i){
		cin>>u[i]>>v[i];
		e[u[i]].push_back(i);e[v[i]].push_back(i);
	}
	if(m&1){cout<<"-1\n";return 0;}
	dfs(1,0);
	for(i=n;i>1;--i){
		int x=dfn[i];
		if(ind[x])ind[x]^=1;else ans[de[x]]^=1,ind[dad[x]]^=1;
	}
	for(i=1;i<=m;++i){
		if(!ans[i])swap(u[i],v[i]);
		cout<<u[i]<<' '<<v[i]<<'\n';
	}
	return 0;
}
### Prompt

In CPP, your task is to solve 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;
const int N=100005;
int T,n,m,i,u[N],v[N];
vector<int>e[N];
int dfn[N],xb,id[N],dad[N],de[N],ans[N];
bool vi[N],ind[N];
void dfs(int x,int fa){
	vi[x]=1;dfn[id[x]=++xb]=x;
	for(int i=0;i<e[x].size();++i){
		int j=e[x][i],y=u[j]^v[j]^x;
		if(ans[j]==-1)ans[j]=x==v[j];
		if(vi[y] && id[y]<id[x] && y!=fa)ind[y]^=1;
		if(!vi[y])dad[y]=x,de[y]=j,dfs(y,x);
	}
}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);
	cin>>n>>m;xb=0;for(i=1;i<=n;++i)e[i].clear();
	memset(ind+1,0,n);memset(ans+1,-1,m<<2);memset(id+1,0,n<<2);memset(vi+1,0,n);
	for(i=1;i<=m;++i){
		cin>>u[i]>>v[i];
		e[u[i]].push_back(i);e[v[i]].push_back(i);
	}
	if(m&1){cout<<"-1\n";return 0;}
	dfs(1,0);
	for(i=n;i>1;--i){
		int x=dfn[i];
		if(ind[x])ind[x]^=1;else ans[de[x]]^=1,ind[dad[x]]^=1;
	}
	for(i=1;i<=m;++i){
		if(!ans[i])swap(u[i],v[i]);
		cout<<u[i]<<' '<<v[i]<<'\n';
	}
	return 0;
}
```