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;
const int maxn=100100;
int n,m,d[maxn],fa[maxn];
inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
vector<pair<int,int> > ans;
vector<int> ver[maxn];
inline void dfs(int x,int fat)
{
	for(int i=0;i<(int)ver[x].size();i++)
	{
		int y=ver[x][i];if(y==fat) continue;dfs(y,x);
		if(d[y]&1) ans.push_back(make_pair(y,x));
		else d[x]++,ans.push_back(make_pair(x,y));
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1,a,b;i<=m;i++)
	{
		scanf("%d%d",&a,&b);
		if(find(a)==find(b)) d[a]++,ans.push_back(make_pair(a,b));
		else ver[a].push_back(b),ver[b].push_back(a),fa[find(a)]=find(b);
	}
	dfs(1,0);
	if(d[1]&1) return puts("-1"),0;
	for(int i=0;i<(int)ans.size();i++) printf("%d %d\n",ans[i].first,ans[i].second);
	return 0;
}