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>
#define pb push_back
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using vi = vector<ll>;
using vvi = vector<vi>;
using pi = pair<ll, ll>;
using vpi = vector<pi>;
const ll mod = 1000000007;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, m, deg[200200];
vvi g;
vi h;
void dfs1(int v = 1) {
	for(auto i : g[v]) {
		if(h[i]>h[v])
			cout << v << " " << i << "\n", deg[v]++;
		else if(h[i]==-1)
			h[i] = h[v]+1, dfs1(i);
	}
}
void dfs2(int v = 1) {
	for(auto i : g[v]) {
		if(h[i]!=-1)
			continue;
		h[i] = h[v]+1, dfs2(i);
		if(deg[i]&1)
			cout << i << " " << v << "\n", deg[i]++;
		else
			cout <<v << " " << i << "\n", deg[v]++;
	}
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> m;
	if(m&1)
		return cout << -1, 0;
	g.resize(n+1);
	for(int f, t, i = 0; i < m; i++) {
		cin >> f >> t;
		g[f].pb(t);
		g[t].pb(f);
	}
	h.assign(n+1, -1), h[1]=0;
	dfs1();
	h.assign(n+1, -1), h[1]=0;
	dfs2();
	return 0;
}