#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000000 + 20 ;
long long inf = 1e18;
long long a[MAXN], ev[MAXN], root[MAXN];
bool vis[MAXN];
vector <int> vec[MAXN];

void dfs(int u, int v) {
	vis[u] = 1;
	root[u] = 1;
	for (int i = 0; i < vec[u].size(); i++)
		if (!vis[vec[u][i]]) {
			dfs(vec[u][i], u);
			if(ev[vec[u][i]] == 0) {
//				cout << "X1 ";
				cout << u << " " << vec[u][i] << endl;
				ev[u] = abs(ev[u] - 1);
			}
			else {
//				cout << "X2 ";
				cout << vec[u][i] << " " << u << endl;
				ev[vec[u][i]] = abs(ev[vec[u][i]] - 1);
			}
		}
		else
			if(root[vec[u][i]] == 0) {
//				cout << "X3 ";
				cout << u << " " << vec[u][i] << endl;
				ev[u] = abs(ev[u] - 1);
			}
			
	root[u] = 0;
}

int main() {
	ios_base::sync_with_stdio(0);cin.tie(0); cout.tie(0);
	int n, m;
	cin >> n >> m;
	
	int u, v;
	for (int i = 0; i < m; i++) {
		cin >> u >> v;
		
		vec[u].push_back(v);
		vec[v].push_back(u);
	}
	
	if (m % 2 == 1)
		return cout << -1, 0;
		
	for(int i = 1; i <= n; i++)
		if(vec[i].size() > 1) {
			dfs(i, 0);
			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 <bits/stdc++.h>
using namespace std;
const int MAXN = 1000000 + 20 ;
long long inf = 1e18;
long long a[MAXN], ev[MAXN], root[MAXN];
bool vis[MAXN];
vector <int> vec[MAXN];

void dfs(int u, int v) {
	vis[u] = 1;
	root[u] = 1;
	for (int i = 0; i < vec[u].size(); i++)
		if (!vis[vec[u][i]]) {
			dfs(vec[u][i], u);
			if(ev[vec[u][i]] == 0) {
//				cout << "X1 ";
				cout << u << " " << vec[u][i] << endl;
				ev[u] = abs(ev[u] - 1);
			}
			else {
//				cout << "X2 ";
				cout << vec[u][i] << " " << u << endl;
				ev[vec[u][i]] = abs(ev[vec[u][i]] - 1);
			}
		}
		else
			if(root[vec[u][i]] == 0) {
//				cout << "X3 ";
				cout << u << " " << vec[u][i] << endl;
				ev[u] = abs(ev[u] - 1);
			}
			
	root[u] = 0;
}

int main() {
	ios_base::sync_with_stdio(0);cin.tie(0); cout.tie(0);
	int n, m;
	cin >> n >> m;
	
	int u, v;
	for (int i = 0; i < m; i++) {
		cin >> u >> v;
		
		vec[u].push_back(v);
		vec[v].push_back(u);
	}
	
	if (m % 2 == 1)
		return cout << -1, 0;
		
	for(int i = 1; i <= n; i++)
		if(vec[i].size() > 1) {
			dfs(i, 0);
			return 0;
		}
}	

	
 
	

	
	
	
```