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

int N;
vector<vector<int>> G, H, des;
int sum[100010], dep[100010];

int dfs(int key){
	sum[key]=0;
	for(int i=0; i<G[key].size(); ++i){
		if(dep[key]<dep[G[key][i]]){
			H[key].emplace_back(G[key][i]);
			if(N<dep[G[key][i]]){
				dep[G[key][i]]=dep[key]+1;
				des[key].emplace_back(dfs(G[key][i])+1);
			}else{
				des[key].emplace_back(1);
			}
			sum[key] += des[key].back();
		}
	}
	return sum[key];
}

int main() {
	int M;
	cin >> N >> M;
	G.resize(N+1);
	for(int i=0; i<M; ++i){
		int A, B;
		cin >> A >> B;
		G[A].emplace_back(B);
		G[B].emplace_back(A);
	}
	if(M%2==1){
		cout << -1 << endl;
		return 0;
	}
	H.resize(N+1);
	des.resize(N+1);
	for(int i=2; i<=N; ++i) dep[i]=N+5;
	dep[1]=0;
	dfs(1);
	for(int i=1; i<=N; ++i){
		for(int j=0; j<H[i].size(); ++j){
			int C=i, D=H[i][j];
			if(des[i][j]%2==0) swap(C, D);
			cout << C << ' ' << D << endl;
		}
	}
	return 0;
}
### Prompt

Your task is to create 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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int N;
vector<vector<int>> G, H, des;
int sum[100010], dep[100010];

int dfs(int key){
	sum[key]=0;
	for(int i=0; i<G[key].size(); ++i){
		if(dep[key]<dep[G[key][i]]){
			H[key].emplace_back(G[key][i]);
			if(N<dep[G[key][i]]){
				dep[G[key][i]]=dep[key]+1;
				des[key].emplace_back(dfs(G[key][i])+1);
			}else{
				des[key].emplace_back(1);
			}
			sum[key] += des[key].back();
		}
	}
	return sum[key];
}

int main() {
	int M;
	cin >> N >> M;
	G.resize(N+1);
	for(int i=0; i<M; ++i){
		int A, B;
		cin >> A >> B;
		G[A].emplace_back(B);
		G[B].emplace_back(A);
	}
	if(M%2==1){
		cout << -1 << endl;
		return 0;
	}
	H.resize(N+1);
	des.resize(N+1);
	for(int i=2; i<=N; ++i) dep[i]=N+5;
	dep[1]=0;
	dfs(1);
	for(int i=1; i<=N; ++i){
		for(int j=0; j<H[i].size(); ++j){
			int C=i, D=H[i][j];
			if(des[i][j]%2==0) swap(C, D);
			cout << C << ' ' << D << endl;
		}
	}
	return 0;
}
```