#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,e;
vector<vector<int> > g,g2;
vector<pair<int,int> > sol;
vector<bool> vis;
set<pair<int,int> > evis;
void DFS(int u,int p){
	vis[u]=true;
	for(auto &v:g[u])
		if(!vis[v]){
			g2[u].push_back(v);
			DFS(v,u);
		}else if(v!=p && evis.insert({min(u,v),max(u,v)}).second)
			sol.push_back({u,v});
}
vector<int> d;
bool calc(int u,int p){
	if(g[u].empty())
		return d[u];
	bool ret=d[u];
	for(auto &v:g[u])
		if(v!=p){
			bool cur=calc(v,u);
			if(cur)
				sol.push_back({v,u});
			else
				sol.push_back({u,v});
			ret^=cur;
		}
	return ret;
}
int main() {
	scanf("%d%d",&n,&e);
	g.resize(n);
	g2.resize(n);
	for(int i=0,a,b;i<e;++i){
		scanf("%d%d",&a,&b);
		--a;--b;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	vis.resize(n);
	DFS(0,-1);
	d.resize(n);
	for(auto &x:sol)
		d[x.first]^=1;
	g.swap(g2);
	int sum=0;
	for(int i=0;i<n;++i){
		d[i]^=g[i].size()&1;
		sum+=d[i];
	}
	if(sum&1){
		puts("-1");
		return 0;
	}
	calc(0,-1);
	for(auto &x:sol)
		printf("%d %d\n",x.first+1,x.second+1);
	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;
typedef long long ll;
int n,e;
vector<vector<int> > g,g2;
vector<pair<int,int> > sol;
vector<bool> vis;
set<pair<int,int> > evis;
void DFS(int u,int p){
	vis[u]=true;
	for(auto &v:g[u])
		if(!vis[v]){
			g2[u].push_back(v);
			DFS(v,u);
		}else if(v!=p && evis.insert({min(u,v),max(u,v)}).second)
			sol.push_back({u,v});
}
vector<int> d;
bool calc(int u,int p){
	if(g[u].empty())
		return d[u];
	bool ret=d[u];
	for(auto &v:g[u])
		if(v!=p){
			bool cur=calc(v,u);
			if(cur)
				sol.push_back({v,u});
			else
				sol.push_back({u,v});
			ret^=cur;
		}
	return ret;
}
int main() {
	scanf("%d%d",&n,&e);
	g.resize(n);
	g2.resize(n);
	for(int i=0,a,b;i<e;++i){
		scanf("%d%d",&a,&b);
		--a;--b;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	vis.resize(n);
	DFS(0,-1);
	d.resize(n);
	for(auto &x:sol)
		d[x.first]^=1;
	g.swap(g2);
	int sum=0;
	for(int i=0;i<n;++i){
		d[i]^=g[i].size()&1;
		sum+=d[i];
	}
	if(sum&1){
		puts("-1");
		return 0;
	}
	calc(0,-1);
	for(auto &x:sol)
		printf("%d %d\n",x.first+1,x.second+1);
	return 0;
}
```