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<bits/stdc++.h>
using namespace std;
vector<pair<int,int> > adj[100005],res;
int n,m,s[100005],dg[100005],visited[100005],dem;
void dfs(int u){
	visited[u]=1;
	s[u]+=dg[u];
	for(int i=0;i<adj[u].size();i++){
		int v=adj[u][i].first;
		if(!visited[v]){
			dfs(v);
			s[u]+=s[v];
			if(s[v]%2==1) swap(res[adj[u][i].second].first,res[adj[u][i].second].second);
		}
	}
}
int main(){
	scanf("%d %d",&n,&m);
	for(int i=0;i<m;i++){
		int u,v;
		scanf("%d %d",&u,&v);
		dg[u]++;
		adj[u].push_back(make_pair(v,i));
		adj[v].push_back(make_pair(u,i));
		res.push_back(make_pair(u,v));
	}
	for(int i=1;i<=n;i++){
		dg[i]%=2;
		if(dg[i]==1) dem++;
	}
	if(dem%2==1) printf("-1");
	else{
		dfs(1);
		for(int i=0;i<res.size();i++){
			printf("%d %d\n",res[i].first,res[i].second);
		}
	}
}