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 N = 1 * 100 * 1000 + 20;

int n, m, x, y;
int arr_p[N], mark[N];

bool outDegree[N];

vector <int> gr[N];

queue <pair<int, int>> ans;

void dfs(int);

int main() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		cin >> x >> y;
		gr[x].push_back(y);
		gr[y].push_back(x);
	}
	if (m % 2 == 1) {
		cout << -1;
		return 0;
	}
	dfs(1);
	while (ans.size() > 0) {
		cout << ans.front().first << " " << ans.front().second;
		cout << (ans.size() > 1? "\n": "");
		ans.pop();
	}
}

void dfs(int a) {
	mark[a] = 1;
	for (int u = 0; u < gr[a].size(); u++) {
		if (gr[a][u] == arr_p[a])
			continue;
		if (mark[gr[a][u]] == 1) {
			outDegree[a] = (outDegree[a] + 1) % 2;
			ans.push({a, gr[a][u]});
		}
		if (mark[gr[a][u]] == 0) {
			arr_p[gr[a][u]] = a;
			dfs(gr[a][u]);
		}
	}
	if (outDegree[a] == 0 && a != 1) {
		outDegree[arr_p[a]] = (outDegree[arr_p[a]] + 1) % 2;
		ans.push({arr_p[a], a});
	}
	if (outDegree[a] == 1 && a != 1)
		ans.push({a, arr_p[a]});
	mark[a] = 2;
}