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 NMAX = 100010;
vector <int> adia[NMAX];
int h[NMAX];
vector <int> fii[NMAX];
vector <pair <int, int>> ans;
bool ok = 1;

void dfs(int nod)
{
	vector <int> sus;
	for (auto i : adia[nod]) {
		if (!h[i]) {
			h[i] = 1 + h[nod];
			dfs(i);
		}
		if (h[i] < h[nod])
			sus.push_back(i);
	}
	while (fii[nod].size() >= 2) {
		int a = fii[nod].back();
		fii[nod].pop_back();
		int b = fii[nod].back();
		fii[nod].pop_back();
		ans.push_back({ nod, a });
		ans.push_back({ nod, b });
	}
	
	if (!fii[nod].empty()) {
		if (sus.empty()) {
			ok = 0;
			return;
		}
		int a = fii[nod][0];
		int b = sus.back();
		sus.pop_back();
		ans.push_back({ nod, a });
		ans.push_back({ nod, b });
	}
	
	for (auto i : sus)
		fii[i].push_back(nod);
}

int main()
{
	int n, m, a, b;
	scanf("%d%d", &n, &m);
	
	while (m--) {
		scanf("%d%d", &a, &b);
		adia[a].push_back(b);
		adia[b].push_back(a);
	}
	
	h[1] = 1;
	dfs(1);
	
	if (!ok) {
		cout << "-1\n";
		return 0;
	}
	
	for (auto i : ans)
		cout << i.first << ' ' << i.second << '\n';
	
	return 0;
}