#include <bits/stdc++.h>

using namespace std;

set<int> g[100005], dg[100005];
bool vis[100005]={};
int out=0;

void dfs(int u, int par){
	vis[u]=1;
	for(auto v:g[u]){
		if(!vis[v])
			dfs(v, u);
	}
	
	int ind=0, la;
	for(auto v:dg[u]){
		if(v!=par){
			if(ind&1){
				cout << u << " " << la << endl;
				cout << u << " " << v << endl;
				dg[la].erase(u);
				dg[v].erase(u);
				out+=2;
			}
			la=v;
			ind++;
		}
	}
	if((((dg[u].size())&1)==0) && (u!=1)){
		cout << u << " " << la << endl;
		cout << u << " " << par << endl;
		out+=2;
		dg[la].erase(u);
		dg[par].erase(u);
	}
}

int main(){
	int n,m; cin >> n >> m;
	for(int i=0,u,v;i<m;i++){
		cin >> u >> v;
		g[u].insert(v);
		g[v].insert(u);
		dg[u].insert(v);
		dg[v].insert(u);
	}
	
	if(m&1){
		cout << -1 << endl;
		return 0;
	}
	dfs(1,-1);
	return 0;
}
### 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;

set<int> g[100005], dg[100005];
bool vis[100005]={};
int out=0;

void dfs(int u, int par){
	vis[u]=1;
	for(auto v:g[u]){
		if(!vis[v])
			dfs(v, u);
	}
	
	int ind=0, la;
	for(auto v:dg[u]){
		if(v!=par){
			if(ind&1){
				cout << u << " " << la << endl;
				cout << u << " " << v << endl;
				dg[la].erase(u);
				dg[v].erase(u);
				out+=2;
			}
			la=v;
			ind++;
		}
	}
	if((((dg[u].size())&1)==0) && (u!=1)){
		cout << u << " " << la << endl;
		cout << u << " " << par << endl;
		out+=2;
		dg[la].erase(u);
		dg[par].erase(u);
	}
}

int main(){
	int n,m; cin >> n >> m;
	for(int i=0,u,v;i<m;i++){
		cin >> u >> v;
		g[u].insert(v);
		g[v].insert(u);
		dg[u].insert(v);
		dg[v].insert(u);
	}
	
	if(m&1){
		cout << -1 << endl;
		return 0;
	}
	dfs(1,-1);
	return 0;
}
```