#include<bits/stdc++.h>
 
using namespace std;
 
#define N 300000
 
int n,m,x,y,xb,p[N],q[N],dep[N];
bool vis[N];
vector<int> g[N];
 
bool dfs(int u,int fa){
	int s=0; vis[u]=1; dep[u]=dep[fa]+1;
	for (auto v:g[u])
		if (!vis[v]){
			s^=dfs(v,u);
		}
		else
			if (dep[v]>dep[u]){
				p[++xb]=u; q[xb]=v;
				s^=1;
			}
	if (s){p[++xb]=u; q[xb]=fa;}
	else{p[++xb]=fa; q[xb]=u;}
	return s^1;
}
 
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;++i){
		scanf("%d%d",&x,&y);
		g[x].push_back(y); g[y].push_back(x);
	}
	if (!dfs(1,0)){puts("-1"); return 0;}
	for (int i=1;i<=m;++i) printf("%d %d\n",p[i],q[i]);
	
	return 0;
}

### 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<bits/stdc++.h>
 
using namespace std;
 
#define N 300000
 
int n,m,x,y,xb,p[N],q[N],dep[N];
bool vis[N];
vector<int> g[N];
 
bool dfs(int u,int fa){
	int s=0; vis[u]=1; dep[u]=dep[fa]+1;
	for (auto v:g[u])
		if (!vis[v]){
			s^=dfs(v,u);
		}
		else
			if (dep[v]>dep[u]){
				p[++xb]=u; q[xb]=v;
				s^=1;
			}
	if (s){p[++xb]=u; q[xb]=fa;}
	else{p[++xb]=fa; q[xb]=u;}
	return s^1;
}
 
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;++i){
		scanf("%d%d",&x,&y);
		g[x].push_back(y); g[y].push_back(x);
	}
	if (!dfs(1,0)){puts("-1"); return 0;}
	for (int i=1;i<=m;++i) printf("%d %d\n",p[i],q[i]);
	
	return 0;
}

```