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:
//ΔAGC035B
#include<iostream>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<bitset>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 111111;
int n,m,a[N],b[N],f[N],c[N];
vector<int> v[N];
int fnd(int x){
	if(f[x]==x)
		return x;
	return f[x]=fnd(f[x]);
}
void dfs(int u,int fa){
	int i,o,x;
	for(i=0;i<v[u].size();i=i+1){
		o=v[u][i];
		x=u^a[o]^b[o];
		if(x==fa)
			continue;
		dfs(x,u);
		if(c[x]){
			if(b[o]==x)
				swap(a[o],b[o]);
		}
		else{
			if(b[o]==u)
				swap(a[o],b[o]);
			c[u]^=1;
		}
	}
}
int main()
{
	int i,x,y;
	scanf("%d%d",&n,&m);
	if(m&1){
		cout<<-1;
		return 0;
	}
	for(i=1;i<=n;i=i+1)
		f[i]=i;
	for(i=1;i<=m;i=i+1){
		scanf("%d%d",&x,&y);
		a[i]=x,b[i]=y;
		if(fnd(x)==fnd(y))
			c[x]^=1;
		else{
			v[x].push_back(i);
			v[y].push_back(i);
			f[fnd(x)]=fnd(y);
		}
	}
	dfs(1,0);
	for(i=1;i<=m;i=i+1)
		printf("%d %d\n",a[i],b[i]);
	return 0;
}