#include <iostream>
#include <vector>
#include <set>
using namespace std;
bool cnt[100005],vis[100005];
vector<int> v[100005];
set<pair<int,int> > ans;
bool dfs(int node,int p)
{
	vis[node]=1;
	bool f=cnt[node];
	for (int u:v[node])
	{
		if (!vis[u])
		f^=dfs(u,node);
	}
	if (node!=1 && f)
	{
		ans.erase({min(node,p),max(node,p)});
		ans.insert({max(node,p),min(node,p)});
	}
	return f;
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	while (m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
		cnt[min(a,b)]^=1;
		ans.insert({min(a,b),max(a,b)});
	}
	if (dfs(1,0))
	printf("-1");
	else
	{
		for (auto p:ans)
		printf("%d %d\n",p.first,p.second);
	}
}
### Prompt

Please create a solution in CPP 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 <set>
using namespace std;
bool cnt[100005],vis[100005];
vector<int> v[100005];
set<pair<int,int> > ans;
bool dfs(int node,int p)
{
	vis[node]=1;
	bool f=cnt[node];
	for (int u:v[node])
	{
		if (!vis[u])
		f^=dfs(u,node);
	}
	if (node!=1 && f)
	{
		ans.erase({min(node,p),max(node,p)});
		ans.insert({max(node,p),min(node,p)});
	}
	return f;
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	while (m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
		cnt[min(a,b)]^=1;
		ans.insert({min(a,b),max(a,b)});
	}
	if (dfs(1,0))
	printf("-1");
	else
	{
		for (auto p:ans)
		printf("%d %d\n",p.first,p.second);
	}
}
```