#include <iostream>
#include <vector>

using namespace std;

int n, m;
vector<int> G[100005];
int deg[100005];
int used[100005];

int id = 1;
void dfs(int v, int p)
{
	used[v] = id++;
	for(int i = 0; i < G[v].size(); i++){
		if(G[v][i] == p) continue;
		if(used[G[v][i]]){
			if(used[G[v][i]] < used[v]){
				deg[v]++;
				cout << v << " " << G[v][i] << endl;
			}
		}
		else dfs(G[v][i], v);
	}
	if(p != -1){
		if(deg[v]%2) cout << v << " " << p << endl;
		else cout << p << " " << v << endl, deg[p]++;
	}
}

int main(void)
{
	cin >> n >> m;
	int u, v;
	for(int i = 1; i <= m; i++){
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	if(m%2){
		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 <iostream>
#include <vector>

using namespace std;

int n, m;
vector<int> G[100005];
int deg[100005];
int used[100005];

int id = 1;
void dfs(int v, int p)
{
	used[v] = id++;
	for(int i = 0; i < G[v].size(); i++){
		if(G[v][i] == p) continue;
		if(used[G[v][i]]){
			if(used[G[v][i]] < used[v]){
				deg[v]++;
				cout << v << " " << G[v][i] << endl;
			}
		}
		else dfs(G[v][i], v);
	}
	if(p != -1){
		if(deg[v]%2) cout << v << " " << p << endl;
		else cout << p << " " << v << endl, deg[p]++;
	}
}

int main(void)
{
	cin >> n >> m;
	int u, v;
	for(int i = 1; i <= m; i++){
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	if(m%2){
		cout << -1 << endl;
		return 0;
	}
	
	dfs(1, -1);
	
	return 0;
}
```