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 = 123456;
int vis[maxn];
vector<int> e[maxn];
vector<pair<int,int> > res;
int dfs(int u, int p){
	vis[u] = 1;
	int s = 0;
	for(int v : e[u]){
		if(v == p) continue;
		if(vis[v] == 2) continue;
		if(vis[v] == 1) s ^= 1, res.push_back({u, v});
		else {
			int vs = dfs(v, u);
			if(vs) res.push_back({v, u});
			else res.push_back({u, v}), s ^= 1;
		}
	}
	vis[u] = 2;
	return s;
}
int main(){
	int n, m; cin >> n >> m;
	for(int i = 0; i < m; i++){
		int u, v; cin >> u >> v;
		e[u].push_back(v);
		e[v].push_back(u);
	}
	if(dfs(1, 0)) cout << -1 << endl;
	else {
		for(auto e : res) cout << e.first << " " << e.second << endl;
	}
	return 0;
}