#include <bits/stdc++.h>
#define ff first
#define ss second
using namespace std;
#define MP make_pair
#define PB push_back
#define ll long long
#define int long long
#define f(i,x,n) for(int i=x;i<n;i++)
#define ld long double
#define mod 1000000007

int n,m;
vector <pair <int,int> > gp[100005];
int va[100005],su;
int vis[100005];
vector <pair <int,int> > ans;
int dfs(int i,int par)
{
	if(vis[i]==1)
		return 0;
	vis[i]=1;
	int vaa=va[i];
	for(auto &j: gp[i])
	{
		int tem=dfs(j.ff,i);
		j.ss^=tem;
		vaa^=tem;
	}
	for(auto &j: gp[i])
	{
		if(j.ff==par)
		{
			j.ss^=vaa;
		}
	}
	return vaa;
}
int32_t main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin>>n>>m;
	f(i,0,m)
	{
		int a,b;
		cin>>a>>b;
		gp[a].push_back({b,1});
		va[a]^=1;
		gp[b].push_back({a,0});
	}
	if(m%2!=0)
	{
		cout<<"-1\n";
		return 0;
	}
	dfs(1,0);
	f(i,1,n+1)
	{
		for(auto j: gp[i])
		{
			if(j.ss==1)
				cout<<i<<" "<<j.ff<<'\n';
		}
	}
	return 0;
}
### Prompt

Please formulate 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>
#define ff first
#define ss second
using namespace std;
#define MP make_pair
#define PB push_back
#define ll long long
#define int long long
#define f(i,x,n) for(int i=x;i<n;i++)
#define ld long double
#define mod 1000000007

int n,m;
vector <pair <int,int> > gp[100005];
int va[100005],su;
int vis[100005];
vector <pair <int,int> > ans;
int dfs(int i,int par)
{
	if(vis[i]==1)
		return 0;
	vis[i]=1;
	int vaa=va[i];
	for(auto &j: gp[i])
	{
		int tem=dfs(j.ff,i);
		j.ss^=tem;
		vaa^=tem;
	}
	for(auto &j: gp[i])
	{
		if(j.ff==par)
		{
			j.ss^=vaa;
		}
	}
	return vaa;
}
int32_t main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin>>n>>m;
	f(i,0,m)
	{
		int a,b;
		cin>>a>>b;
		gp[a].push_back({b,1});
		va[a]^=1;
		gp[b].push_back({a,0});
	}
	if(m%2!=0)
	{
		cout<<"-1\n";
		return 0;
	}
	dfs(1,0);
	f(i,1,n+1)
	{
		for(auto j: gp[i])
		{
			if(j.ss==1)
				cout<<i<<" "<<j.ff<<'\n';
		}
	}
	return 0;
}
```