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;
#define ref(i,x,y)for(int i=x;i<=y;++i)
#define def(i,x,y)for(int i=x;i>=y;--i)
#define pb push_back
#define SZ(x) ((int)x.size())
#define fi first
#define se second
#define mp make_pair
const int N=100010;
int n,m;
vector<int> e[N];
bool vis[N];
int dfn[N],cnt;
int dfs(int la,int x){
	dfn[x]=++cnt;vis[x]=1;
	int count=0;
	ref(i,0,SZ(e[x])-1){
		int y=e[x][i];
		if(vis[y]){
			if(y!=la&&dfn[y]<dfn[x]){
				cout<<x<<" "<<y<<endl;
				count++;
				continue;
			}
			continue;
		}else
		count+=dfs(x,y);
	}
	if(!la)return 0;
	if(count%2==0){
		cout<<la<<" "<<x<<endl;
		return 1;
	}else{
		cout<<x<<" "<<la<<endl;
		return 0;
	}
}
int main(){
	cin>>n>>m;
	if(m%2!=0){
		cout<<"-1"<<endl;
		return 0;
	}
	ref(i,1,m){
		int x,y;cin>>x>>y;
		e[x].pb(y);e[y].pb(x);
	}
	dfs(0,1);
}