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<cstring>
#include<algorithm>
using namespace std;
const int MAXN=1e5+5;
const int MAXM=2e5+5;
int edge,head[MAXN],tail[MAXM],nex[MAXM];
int rd[MAXN],d[MAXN];
int n,m;
void add(int u,int v){
	edge++,nex[edge]=head[u],head[u]=edge,tail[edge]=v;
}
void dfs1(int u){
	for (int e=head[u];e;e=nex[e]){
		int v=tail[e];
		if (d[v]==-1){
			d[v]=d[u]+1; dfs1(v);
		}
		else if (d[v]>d[u]){
			printf("%d %d\n",u,v);
//			cout<<"in"<<endl;
			rd[u]++;
//			cout<<"out"<<endl;
		}
	}
}
void dfs2(int u,int p){
	for (int e=head[u];e;e=nex[e]){
		int v=tail[e];
		if (d[v]==d[u]+1) dfs2(v,u);
	}
//	cout<<u<<" "<<p<<endl;
	if (u==1) return;
	if (rd[u]&1) printf("%d %d\n",u,p);
	else printf("%d %d\n",p,u),rd[p]++;
}
int main(){
	scanf("%d%d",&n,&m);
//	if (m&1) return printf("0\n"),0;
	for (int i=1;i<=m;i++){
		int u,v; scanf("%d%d",&u,&v);
		add(u,v); add(v,u);
	}
	if (m&1) return printf("-1\n"),0;
	for (int i=2;i<=n;i++) d[i]=-1;
	dfs1(1); 
//	for (int i=1;i<=n;i++) printf("id:%d d:%d rd:%d\n",i,d[i],rd[i]);
	dfs2(1,0);
	return 0;
}