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>
#define fi first
#define se second
using namespace std;
const int N = 1e5+3;
vector <pair <int,int> > adj[N];
int deg[N];
bool used[N];
int size[N], dir[N];
int n, m;
pair<int, int> edge[N];

void dfs(int u)
{
	used[u] = true;
	size[u] = deg[u];
	for(auto i:adj[u])
	{
		if(!used[i.fi])
		{
			dfs(i.fi);
			if(size[i.fi] % 2 == 1)
			{
				dir[i.se]^=1;
			}
			size[u] += size[i.fi];
		}
	}
}


int main()
{
	int res = 1;
	cin >> n >> m;
	for(int i=1;i<=m;i++)
	{
		int x, y;
		cin >> x >> y;
		deg[x]^=1;
		adj[x].push_back({y, i});
		adj[y].push_back({x, i});
		edge[i]={x,y};
	}
	
	if(m % 2 == 1)
		return cout <<-1, 0;
	dfs(1);
	for(int i=1;i<=m;i++)
	{
		if(dir[i] == 0)
			cout << edge[i].fi << ' ' << edge[i].se << '\n';	
		else
			cout << edge[i].se << ' ' << edge[i].fi << '\n';
					
	}
	
}