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:
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() { 
  ios::sync_with_stdio(false); cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vector<int>> g(n);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  if (m % 2 == 1) {
    cout << -1 << endl;
    return 0;
  }
  vector<pair<int, int>> ans;
  vector<int> deg(n), vis(n), depth(n);
  function<void(int, int)> dfs = [&] (int cur, int par) {
    vis[cur] = 1;
    for (int child : g[cur]) {
      if (!vis[child]) {
        depth[child] = depth[cur] + 1;
        dfs(child, cur);
      } else {
        if (depth[child] > depth[cur]) {
          ans.emplace_back(cur, child);
          deg[cur]++;
        }
      }
    }
    if (deg[cur] % 2 == 1) {
      ans.emplace_back(cur, par);
      deg[cur]++;
    } else {
      if (par != -1) {
        ans.emplace_back(par, cur);
        deg[par]++;
      }
    }
  };
  dfs(0, -1);
  for (auto i : ans) {
    cout << i.first + 1 << " " << i.second + 1 << '\n';
  }
  return 0;
}