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 ll long long
#define pb push_back
#define mp make_pair
const int N=100050;
vector<int> E[N];
bool use[N];
int disc[N],tid;
vector<pair<int,int>> ans;
void DFS(int u, int p)
{
	disc[u]=++tid;
	vector<int> go;
	for(int v:E[u])
	{
		if(!disc[v])
		{
			DFS(v,u);
			if(!use[v]) go.pb(v);
		}
		else if(disc[v]>disc[u]) go.pb(v);
	}
	if(go.size()&1) go.pb(p),use[u]=1;
	for(int i=0;i<go.size();i++) ans.pb({u,go[i]});
}
int main()
{
	int n,m,u,v;
	scanf("%i %i",&n,&m);
	while(m--) scanf("%i %i",&u,&v),E[u].pb(v),E[v].pb(u);
	DFS(1,0);
	if(use[1]) printf("-1\n");
	else
	{
		for(auto p:ans) printf("%i %i\n",p.first,p.second);
	}
	return 0;
}