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>
#include <set>
#include <utility>
using namespace std;
vector<pair<int,int>> G[100010];
bool used[100010];
set<int> st;
int a[100010] = {},b[100010];
pair<int,int> p[100010];
vector<int> H[100010];
void dfs(int s){
	used[s] = true;
	for(auto x:G[s]){
		int v = x.first;
		if(used[v]) continue;
		st.insert(x.second);
		H[s].push_back(v); H[v].push_back(s);
		b[s]++; b[v]++;
		dfs(v);
	}
}

void dfs2(int s,int p){
	for(int x:H[s]){
		if(x==p) continue;
		dfs2(x,s);
	}
	if(b[s]==1){
		b[s]--; b[p]--;
		if(a[s]&1){
			cout << s + 1 << " " << p + 1 << endl;
			a[s]++;
		}else{
			cout << p + 1 << " " << s + 1 << endl;
			a[p]++;
		}
	}
}
int main(){
	int i,n,m;
	cin >> n >> m;
	if(m&1){
		cout << -1 << endl;
		return 0;
	}
	for(i=0;i<m;i++){
		int a,b; cin >> a >> b; a--; b--;
		G[a].push_back({b,i}); G[b].push_back({a,i});
		p[i] = {a,b};
	}
	for(i=0;i<n;i++) used[i] = false;
	dfs(0);
	for(i=0;i<m;i++){
		if(!st.count(i)){
			cout << p[i].first + 1 << " " << p[i].second + 1 << endl;
			a[p[i].first]++;
		}
	}
	dfs2(0,-1);
}