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;
using lint = long long;
using pi = pair<int, int>;
const int mod = 1e9 + 7;
const int MAXN = 100005;

int n, m;
vector<int> gph[MAXN];
vector<pi> ans;

int vis[MAXN], instk[MAXN];
int prt[MAXN];

int dfs(int x, int p){
	instk[x] = 1;
	vis[x] = 1;
	for(auto &i : gph[x]){
		if(i == p) continue;
		if(!vis[i]){
			int out = dfs(i, x);
			if(out == 0) ans.emplace_back(x, i);
			else ans.emplace_back(i, x);
			prt[x] ^= (1 - out);
		}
		else if(instk[i] && i != p){
			prt[x] ^= 1;
			ans.emplace_back(x, i);
		}
	}
	instk[x] = 0;
	return prt[x];
}

int main(){
	scanf("%d %d",&n,&m);
	for(int i=0; i<m; i++){
		int s, e; scanf("%d %d",&s,&e);
		gph[s].push_back(e);
		gph[e].push_back(s);
	}
	if(dfs(1, 0)){
		puts("-1");
		return 0;
	}
	for(auto &i : ans) printf("%d %d\n", i.first, i.second);
}