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 <set>
#include <map>
#include <vector>
#include <cstdlib>
using namespace std;

const int maxn = 1e5+1;
vector<int> e[maxn];
set<pair<int, int> > visE;
set<int> vis;

void dfs(int u, int fa = -1) {
  vis.insert(u);
  int degu = 0;
  for(auto v : e[u]) {
    if(fa == v) continue;
    if(vis.find(v) == vis.end()) {
      dfs(v, u);
    }
    auto ituv = visE.find(make_pair(u, v));
    auto itvu = visE.find(make_pair(v, u));
    if(ituv != visE.end()) {
      degu++;
    }
    if(ituv == visE.end() && itvu == visE.end()) {
      visE.insert(make_pair(v, u));
    }
  }
  if(fa != -1) {
    if(degu % 2 == 0) {
      visE.insert(make_pair(fa, u));
    }
    else {
      visE.insert(make_pair(u, fa));
    }
  }
}

int main() {
  int N, M, x, y;
  cin>>N>>M;
  for(int i = 0; i < M; i++) {
    cin>>x>>y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(1);
  if(M % 2 == 1) {
    cout<<"-1"<<endl;
  } else {
    for(auto pr: visE) {
      cout<<pr.first<<" "<<pr.second<<endl;
    }
  }
  return 0;
}