#include <iostream>
#include <vector>
using namespace std;

int c[100009];
bool vs[100009], evs[100009];
vector<pair<int, int> > v[100009], ans;

void dfs(int n)
{
	vs[n] = true;
	for (int i = 0; i < v[n].size(); i++) {
		int ei = v[n][i].second;
		if (evs[ei]) continue;
		evs[ei] = true;
		int tn = v[n][i].first;
		if (vs[tn]) {
			ans.push_back(make_pair(n, tn));
			c[n]++;
			continue;
		}
		dfs(tn);
		if (c[tn] % 2) 
			ans.push_back(make_pair(tn, n));
		else {
			ans.push_back(make_pair(n, tn));
			c[n]++;
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, m; cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b; cin >> a >> b;
		v[a].push_back(make_pair(b, i)); v[b].push_back(make_pair(a, i));
	}
	if (m % 2) {
		cout << -1 << '\n';
		return 0;
	}
	dfs(1);
	for (int i = 0; i < m; i++)
		cout << ans[i].first << " " << ans[i].second << '\n';
	return 0;
}
### Prompt

Develop a solution in Cpp 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 c[100009];
bool vs[100009], evs[100009];
vector<pair<int, int> > v[100009], ans;

void dfs(int n)
{
	vs[n] = true;
	for (int i = 0; i < v[n].size(); i++) {
		int ei = v[n][i].second;
		if (evs[ei]) continue;
		evs[ei] = true;
		int tn = v[n][i].first;
		if (vs[tn]) {
			ans.push_back(make_pair(n, tn));
			c[n]++;
			continue;
		}
		dfs(tn);
		if (c[tn] % 2) 
			ans.push_back(make_pair(tn, n));
		else {
			ans.push_back(make_pair(n, tn));
			c[n]++;
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, m; cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b; cin >> a >> b;
		v[a].push_back(make_pair(b, i)); v[b].push_back(make_pair(a, i));
	}
	if (m % 2) {
		cout << -1 << '\n';
		return 0;
	}
	dfs(1);
	for (int i = 0; i < m; i++)
		cout << ans[i].first << " " << ans[i].second << '\n';
	return 0;
}
```