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<stdio.h>
#include<vector>
using namespace std;
vector< vector<int> > adj;
vector< pair<int,int> > res;
int check[1000000],num,v,out[1000000],n,m,u;
void DFS(int pos,int par){
	check[pos]=++num;
	for(int i=0;i<adj[pos].size();i++){
		int nxt=adj[pos][i];
		if(check[nxt]==0){
			DFS(nxt,pos);
			if(out[nxt]%2==1){
				res.push_back(make_pair(nxt,pos));
				out[nxt]++;
			}else{
				res.push_back(make_pair(pos,nxt));
				out[pos]++;
			}
		}else if(nxt!=par&&check[nxt]<check[pos]){
			out[pos]++;
			res.push_back(make_pair(pos,nxt));
		}
	}
}
int main(){
	scanf("%d %d", &n,&m);
	adj.resize(n+1);
	for(int i=1;i<=m;i++){
		scanf("%d %d", &u,&v);
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	if(m%2==1){
		printf("-1");
		return 0;
	}
	DFS(1,0);
	for(int i=0;i<res.size();i++) printf("%d %d\n", res[i].first,res[i].second);
}