#include <bits/stdc++.h>

using namespace std;

typedef pair <int, int> pii;

vector <int> T[101010];
vector <pii> A;
int D[101010];
bool chk[101010];
int n, m;

bool dfs(int p, int r)
{
	int x = 0;
	
	D[p] = D[r] + 1;
	chk[p] = 1;
	
	for(int &t: T[p]){
		if(!chk[t]){
			if(dfs(t, p)) x ++;
		}
		else if(t != r){
			if(D[t] < D[p]){
				A.emplace_back(p, t);
				x ++;
			}
		}
	}
	
	if(x & 1) A.emplace_back(p, r);
	else{
		if(r) A.emplace_back(r, p);
		return 1;
	}
	
	return 0;
}

int main()
{
	int i, u, v;
	
	scanf("%d%d", &n, &m);
	
	for(i=0; i<m; i++){
		scanf("%d%d", &u, &v);
		T[u].push_back(v);
		T[v].push_back(u);
	}
	
	if(!dfs(1, 0)) printf("-1\n");
	else{
		for(pii &t: A){
			printf("%d %d\n", t.first, t.second);
		}
	}
	
	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;

typedef pair <int, int> pii;

vector <int> T[101010];
vector <pii> A;
int D[101010];
bool chk[101010];
int n, m;

bool dfs(int p, int r)
{
	int x = 0;
	
	D[p] = D[r] + 1;
	chk[p] = 1;
	
	for(int &t: T[p]){
		if(!chk[t]){
			if(dfs(t, p)) x ++;
		}
		else if(t != r){
			if(D[t] < D[p]){
				A.emplace_back(p, t);
				x ++;
			}
		}
	}
	
	if(x & 1) A.emplace_back(p, r);
	else{
		if(r) A.emplace_back(r, p);
		return 1;
	}
	
	return 0;
}

int main()
{
	int i, u, v;
	
	scanf("%d%d", &n, &m);
	
	for(i=0; i<m; i++){
		scanf("%d%d", &u, &v);
		T[u].push_back(v);
		T[v].push_back(u);
	}
	
	if(!dfs(1, 0)) printf("-1\n");
	else{
		for(pii &t: A){
			printf("%d %d\n", t.first, t.second);
		}
	}
	
	return 0;
}
```