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 ll = long long;
using P = pair<int,int>;
const ll inf = numeric_limits<ll>::max()/3;
const ll mod = 1e9 + 7;

vector<vector<P>> G;
vector<P> E;
int odd[100010];
int rev[100010];
int visit[100010];

int dfs(int n){
  visit[n] = 1;
  int res = odd[n];
  for(auto p : G[n]){
    if(!visit[p.first]){
      if(dfs(p.first)){
        rev[p.second] = 1;
        res = !res;
      }
    }
  }
  return res;
}

int main(void){
  int n,m;
  scanf("%d %d", &n, &m);
  G = vector<vector<P>>(n);
  E = vector<P>(m);
  for(int i = 0;i < m;++i){
    int u,v;
    scanf("%d %d", &u, &v);
    --u;--v;
    ++odd[u];
    E[i] = P(u,v);
    G[u].push_back(P(v,i));
    G[v].push_back(P(u,i));
  }
  for(int i = 0;i < n;++i){
    odd[i] %= 2;
  }
  fill(rev, rev+100010, 0);
  fill(visit, visit+100010, 0);
  if(dfs(0)){
    puts("-1");
  }else{
    for(int i = 0;i < m;++i){
      if(rev[i]){
        printf("%d %d\n", E[i].second+1, E[i].first+1);
      }else{
        printf("%d %d\n", E[i].first+1, E[i].second+1);
      }
    }
  }
  return 0;
}