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:
/*Lucky_Glass*/
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5;
vector<int> lnk[N+3];
int n,m,nidx;
int du[N+3],dep[N+5];
int idx[N+3];
vector<pair<int,int> >ans;
int DFS(int u,int depth,int pre){
	idx[u]=++nidx;
	dep[u]=depth;
	int ret=0;
	for(auto v : lnk[u]){
		if(v==pre) continue;
		if(dep[v]){
			if(dep[v]<dep[u]) ret++,ans.push_back(make_pair(u,v));
			continue;
		}
		int retv=DFS(v,depth+1,u);
		if(retv%2==0) ret++,ans.push_back(make_pair(u,v));
		else ans.push_back(make_pair(v,u));
	}
	return ret;
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		int u,v;scanf("%d%d",&u,&v);
		lnk[u].push_back(v);
		lnk[v].push_back(u);
		du[u]++;du[v]++;
	}
	int res=DFS(1,1,0);
	if(res%2) printf("-1\n");
	else{
		for(int i=0;i<ans.size();i++)
			printf("%d %d\n",ans[i].first,ans[i].second);
	}
	return 0;
}