#include<bits/stdc++.h>

using namespace std;

int n, m;
struct edge{
	int u, v, flip;
};
vector<edge> e;
vector<vector<int> > g;
vector<int> vis, f;
int dfs(int cur, int eid){
	vis[cur] = 1;
	for(int u: g[cur]){
		int next = e[u].u + e[u].v - cur;
		if(vis[next] == 1) continue;
		f[cur] ^= dfs(next, u);
	}
	if(f[cur]){
		if(eid != -1){
			assert(eid >= 0 && eid < e.size());
			e[eid].flip ^= 1;
		}
	}
	return f[cur];
}

int main(){
	ios::sync_with_stdio(false); cin.tie(0);
	cin>>n>>m;
	if(m%2 != 0){
		cout<<-1<<endl;
		return 0;
	}

	e.resize(m);
	g.resize(n);
	for(int i = 0; i < m; i++){
		cin>>e[i].u>>e[i].v; //egde is right now going from u to v
		e[i].u--; e[i].v--;
		e[i].flip = 0;
		g[e[i].u].push_back(i);
		g[e[i].v].push_back(i);
	}
	vis.resize(n);
	f.resize(n);
	for(int i = 0; i < n; i++){
		f[i] = 0;
		for(int k: g[i]){
			if(e[k].u == i){
				f[i] ^= 1;
			}
		}
	}
	dfs(0, -1);
	for(auto ed: e){
		if(ed.flip){
			swap(ed.u, ed.v);
		}
		cout<<ed.u+1<<" "<<ed.v+1<<endl;
	}
}
### 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>

using namespace std;

int n, m;
struct edge{
	int u, v, flip;
};
vector<edge> e;
vector<vector<int> > g;
vector<int> vis, f;
int dfs(int cur, int eid){
	vis[cur] = 1;
	for(int u: g[cur]){
		int next = e[u].u + e[u].v - cur;
		if(vis[next] == 1) continue;
		f[cur] ^= dfs(next, u);
	}
	if(f[cur]){
		if(eid != -1){
			assert(eid >= 0 && eid < e.size());
			e[eid].flip ^= 1;
		}
	}
	return f[cur];
}

int main(){
	ios::sync_with_stdio(false); cin.tie(0);
	cin>>n>>m;
	if(m%2 != 0){
		cout<<-1<<endl;
		return 0;
	}

	e.resize(m);
	g.resize(n);
	for(int i = 0; i < m; i++){
		cin>>e[i].u>>e[i].v; //egde is right now going from u to v
		e[i].u--; e[i].v--;
		e[i].flip = 0;
		g[e[i].u].push_back(i);
		g[e[i].v].push_back(i);
	}
	vis.resize(n);
	f.resize(n);
	for(int i = 0; i < n; i++){
		f[i] = 0;
		for(int k: g[i]){
			if(e[k].u == i){
				f[i] ^= 1;
			}
		}
	}
	dfs(0, -1);
	for(auto ed: e){
		if(ed.flip){
			swap(ed.u, ed.v);
		}
		cout<<ed.u+1<<" "<<ed.v+1<<endl;
	}
}
```