#include <bits/stdc++.h>

using namespace std;

int 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 x, y;
    cin >> x >> y;
    --x; --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  if (m % 2 == 1) {
    cout << -1 << '\n';
    return 0;
  }
  vector<pair<int, int>> res;
  vector<int> deg(n, 0);
  vector<int> depth(n, -1);
  function<void(int, int)> Dfs = [&](int v, int pr) {
    for (int u : g[v]) {
      if (depth[u] == -1) {
        depth[u] = depth[v] + 1;
        Dfs(u, v);
      } else {
        if (depth[u] > depth[v]) {
          res.emplace_back(v, u);
          ++deg[v];
        }
      }
    }
    if (deg[v] % 2 == 1) {

      res.emplace_back(v, pr);
      ++deg[v];
    } else {
      if (pr != -1) {
        res.emplace_back(pr, v);
        ++deg[pr];
      }
    }
  };
  depth[0] = 0;
  Dfs(0, -1);
  for (auto& p : res) {
    cout << p.first + 1 << " " << p.second + 1 << '\n';
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>

using namespace std;

int 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 x, y;
    cin >> x >> y;
    --x; --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  if (m % 2 == 1) {
    cout << -1 << '\n';
    return 0;
  }
  vector<pair<int, int>> res;
  vector<int> deg(n, 0);
  vector<int> depth(n, -1);
  function<void(int, int)> Dfs = [&](int v, int pr) {
    for (int u : g[v]) {
      if (depth[u] == -1) {
        depth[u] = depth[v] + 1;
        Dfs(u, v);
      } else {
        if (depth[u] > depth[v]) {
          res.emplace_back(v, u);
          ++deg[v];
        }
      }
    }
    if (deg[v] % 2 == 1) {

      res.emplace_back(v, pr);
      ++deg[v];
    } else {
      if (pr != -1) {
        res.emplace_back(pr, v);
        ++deg[pr];
      }
    }
  };
  depth[0] = 0;
  Dfs(0, -1);
  for (auto& p : res) {
    cout << p.first + 1 << " " << p.second + 1 << '\n';
  }
  return 0;
}

```