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<bits/stdc++.h>
#define ll long long
#define   ull unsigned long long 
#define mp make_pair
#define pb push_back
const int MAX=2e5+9;
const long long mod=1e9+7;
using namespace std;
vector<int>adj[MAX];
int d[MAX];
int deg[MAX];
void dfs(int x, int y){
	d[x]=y;
	for(int i=0;i<adj[x].size();i++){
		if(d[adj[x][i]]==0){
			dfs(adj[x][i] , y+1);
			if(deg[adj[x][i]]%2==0){
				deg[x]++;
				cout<<x+1<<" "<<adj[x][i]+1<<"\n";
			}
			else{
				deg[adj[x][i]]++;
				cout<<adj[x][i]+1<<" "<<x+1<<"\n";
			}
		}
		else if(d[adj[x][i]]>d[x]){
			deg[x]++;
			cout<<x+1<<" "<<adj[x][i]+1<<"\n";
		}

	}
}

int main(){
	int n , m;
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int x, y;
		cin>>x>>y;
		x--;
		y--;
		adj[x].push_back(y);
		adj[y].push_back(x);
	}
	if(m%2){
		cout<<-1;
	}
	else
	{
		dfs(0, 1);
	}
}