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

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
int n,m;
int f[200010],val[200010],u[200010],v[200010];
vector<int>vec[200010];
inline int read(){
	int ans=0,k=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')k=-1;c=getchar();}
	while(c>='0'&&c<='9'){ans=(ans<<3)+(ans<<1)+(c-'0');c=getchar();}
	return ans*k;
}
int find(int x){
	return f[x]==x?x:f[x]=find(f[x]);
}
bool dfs(int u,int fa){
	int siz=vec[u].size();
	for(register int i=0;i<siz;i++){
		if(vec[u][i]==fa)continue;
		if(dfs(vec[u][i],u))val[u]^=1;
	}
	if(u!=1){
		if(val[u]&1){
			printf("%d %d\n",u,fa);
			return false;
		}
		else{
			printf("%d %d\n",fa,u);
			return true;
		}
	}
}
int main(){
	n=read();m=read();
	if(m&1){
		printf("-1");
		return 0;
	}
	for(register int i=1;i<=n;i++)f[i]=i;
	for(register int i=1;i<=m;i++){
		u[i]=read();v[i]=read();
		if(find(u[i])!=find(v[i])){
			vec[u[i]].push_back(v[i]);
			vec[v[i]].push_back(u[i]);
			f[f[u[i]]]=f[v[i]];
		}
		else{
			val[u[i]]^=1;
			printf("%d %d\n",u[i],v[i]);
		}
	}
	dfs(1,0);
	return 0;
}