#include <bits/stdc++.h>
using namespace std;
int n,m;
int deg[200005];
int dir[200005],depth[200005];
vector<int> g[200005];
vector<pair<int,int> > ans;
void dfs(int x,int fa){
    for(int i=0;i<g[x].size();i++){
    	int y=g[x][i];
		if(y==fa)	continue;
		if(!depth[y]){
            depth[y]=depth[x]+1;
            dfs(y,x);
        }
        else if(depth[y]<depth[x]){
            deg[x]^=1;
            ans.push_back(make_pair(x,g[x][i]));
        }
    }
    if(deg[x]&1){
        deg[x]^=1;
        ans.push_back(make_pair(x,fa));
    }
    else if(fa){
        deg[fa]^=1;
        ans.push_back(make_pair(fa,x));
    }
}
int main(){
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int u,v;cin>>u>>v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	if(m&1)	return puts("-1"),0;
	depth[1]=1;
	dfs(1,0);
	for(int i=0;i<ans.size();i++)
		cout<<ans[i].first<<" "<<ans[i].second<<endl;
	return 0;
}
### Prompt

Please formulate 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;
int n,m;
int deg[200005];
int dir[200005],depth[200005];
vector<int> g[200005];
vector<pair<int,int> > ans;
void dfs(int x,int fa){
    for(int i=0;i<g[x].size();i++){
    	int y=g[x][i];
		if(y==fa)	continue;
		if(!depth[y]){
            depth[y]=depth[x]+1;
            dfs(y,x);
        }
        else if(depth[y]<depth[x]){
            deg[x]^=1;
            ans.push_back(make_pair(x,g[x][i]));
        }
    }
    if(deg[x]&1){
        deg[x]^=1;
        ans.push_back(make_pair(x,fa));
    }
    else if(fa){
        deg[fa]^=1;
        ans.push_back(make_pair(fa,x));
    }
}
int main(){
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int u,v;cin>>u>>v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	if(m&1)	return puts("-1"),0;
	depth[1]=1;
	dfs(1,0);
	for(int i=0;i<ans.size();i++)
		cout<<ans[i].first<<" "<<ans[i].second<<endl;
	return 0;
}
```