#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000 * 100 + 20;
vector <int> gr[maxn], ans[maxn];
int n, m, vis[maxn];
void dfs(int v, int u) {
	vis[v] = 1;
	for (int i: gr[v]) {
		if (!vis[i]) {
			dfs(i , v);
			if (ans[i].size() % 2)
				ans[i].push_back(v);
			else
				ans[v].push_back(i);
			
		}
		else if (i != u) {
			if (i > v)
				ans[v].push_back(i);
		}
	}
}
int main() {
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		gr[a].push_back(b);
		gr[b].push_back(a);
	}
	
	if (m % 2)
		cout << -1 << endl;
	else {
		dfs(1, 0);
	}
	for (int i = 1; i <= n; i++)
		for (int j: ans[i])
			cout << i << " " << j << endl;
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int maxn = 1000 * 100 + 20;
vector <int> gr[maxn], ans[maxn];
int n, m, vis[maxn];
void dfs(int v, int u) {
	vis[v] = 1;
	for (int i: gr[v]) {
		if (!vis[i]) {
			dfs(i , v);
			if (ans[i].size() % 2)
				ans[i].push_back(v);
			else
				ans[v].push_back(i);
			
		}
		else if (i != u) {
			if (i > v)
				ans[v].push_back(i);
		}
	}
}
int main() {
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		gr[a].push_back(b);
		gr[b].push_back(a);
	}
	
	if (m % 2)
		cout << -1 << endl;
	else {
		dfs(1, 0);
	}
	for (int i = 1; i <= n; i++)
		for (int j: ans[i])
			cout << i << " " << j << endl;
	return 0;
}
```