#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define rrep(i,a,b) for(int i=(a);i>=(b);--i)
#define PII pair<int,int>
#define MP make_pair
#define PB push_back
typedef long long LL;
const LL MOD = 998244353;
const int N = 100010;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n,m,x,y,out[N],a[N],v[N],cnt=0;
set<int> g[N];

void process(int x){
	for(auto u:g[x]){
		if(out[x]%2){
			++out[x];
			cout<<x<<" "<<u<<endl;
		}else{
			++out[u];
			cout<<u<<" "<<x<<endl;
		}
		g[u].erase(x);
	}
	g[x].clear();
}

void dfs(int x){
	v[x]=true;
	a[++cnt]=x;
	for(auto u:g[x])if(!v[u])dfs(u);
}

int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin>>n>>m;
	if(m%2){
		cout<<-1<<endl;
		return 0;
	}
	rep(i,1,m)cin>>x>>y,g[x].insert(y),g[y].insert(x);
	
	memset(v,0,sizeof v);
	dfs(1);
	rrep(i,n,1){
		process(a[i]);
	}
	return 0;
}
### Prompt

Create a solution in cpp for 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;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define rrep(i,a,b) for(int i=(a);i>=(b);--i)
#define PII pair<int,int>
#define MP make_pair
#define PB push_back
typedef long long LL;
const LL MOD = 998244353;
const int N = 100010;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n,m,x,y,out[N],a[N],v[N],cnt=0;
set<int> g[N];

void process(int x){
	for(auto u:g[x]){
		if(out[x]%2){
			++out[x];
			cout<<x<<" "<<u<<endl;
		}else{
			++out[u];
			cout<<u<<" "<<x<<endl;
		}
		g[u].erase(x);
	}
	g[x].clear();
}

void dfs(int x){
	v[x]=true;
	a[++cnt]=x;
	for(auto u:g[x])if(!v[u])dfs(u);
}

int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin>>n>>m;
	if(m%2){
		cout<<-1<<endl;
		return 0;
	}
	rep(i,1,m)cin>>x>>y,g[x].insert(y),g[y].insert(x);
	
	memset(v,0,sizeof v);
	dfs(1);
	rrep(i,n,1){
		process(a[i]);
	}
	return 0;
}
```