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 <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
vector<vector<int> > g;
vector<pair<int,int> > res;
vector<int> V;
int dfs(int a) {
	V[a] = 1;
	int c = 0;
	for (int i : g[a]) {
		if (V[i] == 0) {
			int t = dfs(i);
			if (t % 2 == 0) {
				res.emplace_back(a, i);
			}
			else {
				res.emplace_back(i, a);
			}
			c += t + 1;
		}
		else {
			if (V[i] == 2) {
				res.emplace_back(a, i);
				c++;
			}
		}
	}
	V[a] = 2;
	return c;
}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, M;
	cin >> N >> M;
	g.resize(N);
	int a, b;
	for (int i = 0; i < M; i++) {
		cin >> a >> b; a--; b--;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	if (M % 2 == 1) {
		cout << -1 << endl;
		return 0;
	}
	V.resize(N, 0);
	dfs(0);


	for (int i = 0; i < res.size(); i++) {
		cout << res[i].first + 1 << " " << res[i].second + 1  << endl;
	}
}