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 = 100005;
int n, m, x, y;
vector<int> v[N];
vector<pair<int, int> > fp;
bool bo[N], st[N], sh[N];

void sgd(int dg, int op) {
	bo[dg] = sh[dg] = 1;
	for(int sw : v[dg]) {
		if(!bo[sw]) {
			sgd(sw, dg);
		} else {
			if(!sh[sw]) fp.push_back({dg, sw}), st[dg] ^= 1;
		}
	}
	if(dg ^ 1) {
		if(st[dg]) {
			fp.push_back({dg, op});
			st[dg] = 0;
		} else {
			fp.push_back({op, dg});
			st[op] ^= 1;
		}
	}
	sh[dg] = 0;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n >> m;
	if(m & 1)
		return cout << -1 << endl, 0;
	for(int i = 1; i <= m; ++i) {
		cin >> x >> y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	sgd(1, 0);
	for(pair<int, int> a : fp) {
		cout << a.first << " " << a.second << "\n";
	}
	cout << flush;
	return 0;
}