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>

using namespace std;

int n, m;
vector<pair<int,int>> e;
vector<pair<int,int>> v[101010];
int d[101010];
int z[101010];
int b[101010];

int dfs(int x) {
	z[x] = 1;
	int r = 0;
	for (auto u : v[x]) {
		if (z[u.first]) continue;
		int h = dfs(u.first);
		if (h) b[u.second] = 1-b[u.second];
		r += h;
	}
	if (d[x]%2 == 1) r++;
	return r%2;
}

int main() {
	cin >> n >> m;
	if (m%2 == 1) {
		cout << "-1\n";
		return 0;
	}
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		e.push_back({a,b});
		d[a]++;
		v[a].push_back({b,i});
		v[b].push_back({a,i});
	}
	dfs(1);
	for (int i = 0; i < m; i++) {
		if (b[i]) swap(e[i].first,e[i].second);
		cout << e[i].first << " " << e[i].second << "\n";
	}
}