#include<bits/stdc++.h>

#define debug(x) cerr << #x << ": " << x << endl
#define debugArray(x,n) for(long long hoge = 0; (hoge) < (n); ++ (hoge)) cerr << #x << "[" << hoge << "]: " << x[hoge] << endl
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vll;
const ll INF = LLONG_MAX/2;
const ll MOD = 1e9+7;

ll N,M;
vll g[100010];
ll dep[100010];
ll dfs(ll v,ll p,ll d){
  if(dep[v])return 0;
  dep[v]=d;
  ll cnt=0;
  for(ll u:g[v])if(u!=p){
    if(dep[u]&&dep[u]<dep[v]){
      cout<<v+1<<" "<<u+1<<endl;
      cnt++;
    }else{
      cnt+=dfs(u,v,d+1);
    }
  }
  if(p>=0){
    if(cnt&1){
      cout<<v+1<<" "<<p+1<<endl;
      return 0;
    }else{
      cout<<p+1<<" "<<v+1<<endl;
      return 1;
    }
  }
  return 0;
}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin>>N>>M;
  for(ll i=0;i<M;i++){
    ll A,B;cin>>A>>B;A--;B--;
    g[A].push_back(B);
    g[B].push_back(A);
  }
  if(M&1)cout<<-1<<endl;
  else dfs(0,-1,1);
  return 0;
}

### Prompt

Create a solution in cpp for the following 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

### Response

```cpp
#include<bits/stdc++.h>

#define debug(x) cerr << #x << ": " << x << endl
#define debugArray(x,n) for(long long hoge = 0; (hoge) < (n); ++ (hoge)) cerr << #x << "[" << hoge << "]: " << x[hoge] << endl
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vll;
const ll INF = LLONG_MAX/2;
const ll MOD = 1e9+7;

ll N,M;
vll g[100010];
ll dep[100010];
ll dfs(ll v,ll p,ll d){
  if(dep[v])return 0;
  dep[v]=d;
  ll cnt=0;
  for(ll u:g[v])if(u!=p){
    if(dep[u]&&dep[u]<dep[v]){
      cout<<v+1<<" "<<u+1<<endl;
      cnt++;
    }else{
      cnt+=dfs(u,v,d+1);
    }
  }
  if(p>=0){
    if(cnt&1){
      cout<<v+1<<" "<<p+1<<endl;
      return 0;
    }else{
      cout<<p+1<<" "<<v+1<<endl;
      return 1;
    }
  }
  return 0;
}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin>>N>>M;
  for(ll i=0;i<M;i++){
    ll A,B;cin>>A>>B;A--;B--;
    g[A].push_back(B);
    g[B].push_back(A);
  }
  if(M&1)cout<<-1<<endl;
  else dfs(0,-1,1);
  return 0;
}

```