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;
#define int long long
const int MAX = 1000000;
const int MOD = 1000000007;
vector<pair<int,int>> hen[100000];
int mk[2][100000];
bool used[100000]={},tdd[100000]={};
void dfs(int a){

  used[a]=true;
  for(int i=0;i<hen[a].size();i++){
 
    if(used[hen[a][i].first]){

      continue;
    }
    if(tdd[a]){
     swap(mk[0][hen[a][i].second],mk[1][hen[a][i].second]);
      tdd[a]=!tdd[a];
      tdd[hen[a][i].first]=!tdd[hen[a][i].first];
    }
    dfs(hen[a][i].first);
    if(tdd[hen[a][i].first]){
      swap(mk[0][hen[a][i].second],mk[1][hen[a][i].second]);
      tdd[a]=!tdd[a];
      tdd[hen[a][i].first]=!tdd[hen[a][i].first];
    }
  }
}
signed main() {
int n,m;
cin>>n>>m;
int A,b;
for(int i=0;i<m;i++){
  cin>>A>>b;A--;b--;
  mk[0][i]=A;mk[1][i]=b;
  hen[A].push_back(make_pair(b,i));
  hen[b].push_back(make_pair(A,i));
  tdd[A]=!tdd[A];
}

dfs(0);
if(tdd[0])cout<<-1;
else{
  for(int i=0;i<m;i++)cout<<mk[0][i]+1<<' '<<mk[1][i]+1<<endl;
}

}