#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> G;
vector<int> edge_origin, passed_order;
int order(0);

void solve(int node, int parent){
	if(passed_order[node]) return ;
	passed_order[node] = ++order;
	for(int neighbor : G[node]){
		if(neighbor == parent) continue;
		if(passed_order[neighbor]){
			if(passed_order[node] > passed_order[neighbor]){
				edge_origin[node]++;
				cout<<node<<" "<<neighbor<<endl;
			}
		} else {
			solve(neighbor , node);
		}
	}
	if(parent == -1) return;
	if(edge_origin[node]%2){
		cout << node << " " << parent << endl;
		edge_origin[node]++;
	} else {
		cout<<parent<<" "<<node<<endl;
		edge_origin[parent]++;
	}
}
int main(){	
	int N, M;
	cin >> N >> M;
    if(M%2 != 0){
		cout<< -1 <<endl;
		return 0;
	}
    G.resize(N+1);
    edge_origin.resize(N+1);
	passed_order.resize(N+1);
	int a, b;
	for(int i(0); i < M; ++i){
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	solve(1,-1);
	return 0;
}
### Prompt

Generate 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;
vector<vector<int>> G;
vector<int> edge_origin, passed_order;
int order(0);

void solve(int node, int parent){
	if(passed_order[node]) return ;
	passed_order[node] = ++order;
	for(int neighbor : G[node]){
		if(neighbor == parent) continue;
		if(passed_order[neighbor]){
			if(passed_order[node] > passed_order[neighbor]){
				edge_origin[node]++;
				cout<<node<<" "<<neighbor<<endl;
			}
		} else {
			solve(neighbor , node);
		}
	}
	if(parent == -1) return;
	if(edge_origin[node]%2){
		cout << node << " " << parent << endl;
		edge_origin[node]++;
	} else {
		cout<<parent<<" "<<node<<endl;
		edge_origin[parent]++;
	}
}
int main(){	
	int N, M;
	cin >> N >> M;
    if(M%2 != 0){
		cout<< -1 <<endl;
		return 0;
	}
    G.resize(N+1);
    edge_origin.resize(N+1);
	passed_order.resize(N+1);
	int a, b;
	for(int i(0); i < M; ++i){
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	solve(1,-1);
	return 0;
}
```