#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int main() {
	int N, M;
	cin >> N >> M;
	vector<vector<int> > G(N);
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b; --a, --b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	if (M % 2 == 1) {
		cout << -1 << endl;
	}
	else {
		vector<bool> vis(N, false);
		vector<int> out(N), depth(N);
		function<void(int, int)> dfs = [&](int pos, int pre) {
			vis[pos] = true;
			for (int i : G[pos]) {
				if (vis[i]) {
					if (depth[i] < depth[pos] - 1) {
						cout << i + 1 << ' ' << pos + 1 << endl;
						++out[i];
					}
				}
				else {
					depth[i] = depth[pos] + 1;
					dfs(i, pos);
				}
			}
			if (pre != -1) {
				if (out[pos] % 2 == 1) {
					cout << pos + 1 << ' ' << pre + 1 << endl;
					++out[pos];
				}
				else {
					cout << pre + 1 << ' ' << pos + 1 << endl;
					++out[pre];
				}
			}
		};
		dfs(0, -1);
	}
	return 0;
}
### Prompt

Your challenge is to write 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 <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int main() {
	int N, M;
	cin >> N >> M;
	vector<vector<int> > G(N);
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b; --a, --b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	if (M % 2 == 1) {
		cout << -1 << endl;
	}
	else {
		vector<bool> vis(N, false);
		vector<int> out(N), depth(N);
		function<void(int, int)> dfs = [&](int pos, int pre) {
			vis[pos] = true;
			for (int i : G[pos]) {
				if (vis[i]) {
					if (depth[i] < depth[pos] - 1) {
						cout << i + 1 << ' ' << pos + 1 << endl;
						++out[i];
					}
				}
				else {
					depth[i] = depth[pos] + 1;
					dfs(i, pos);
				}
			}
			if (pre != -1) {
				if (out[pos] % 2 == 1) {
					cout << pos + 1 << ' ' << pre + 1 << endl;
					++out[pos];
				}
				else {
					cout << pre + 1 << ' ' << pos + 1 << endl;
					++out[pre];
				}
			}
		};
		dfs(0, -1);
	}
	return 0;
}
```