#include <bits/stdc++.h>
using namespace std;
typedef long long LL;


const int MAXN = 1e5 + 10;
vector<int> g[MAXN];
bool vis[MAXN] = {0};
bool inst[MAXN] = {0};
vector<pair<int, int>> ans;

bool dfs(int u, int p = -1) {
  vis[u] = 1;
  inst[u] = 1;
  int outdeg = 0;
  for (int v : g[u]) if (v != p) {
    if (not vis[v]) {
      if (dfs(v, u)) ++outdeg, ans.emplace_back(u, v);
    }
    else if (inst[v]) ++outdeg, ans.emplace_back(u, v);
  }
  if (outdeg&1) ans.emplace_back(u, p);
  inst[u] = 0;
  return outdeg % 2 == 0;
}

int main() {
  ios_base::sync_with_stdio(false); cin.tie(0);
  int n, m; cin >> n >> m;
  if (m % 2) cout << -1 << '\n', exit(0);
  for (int i = 0; i < m; ++i) {
    int u, v; cin >> u >> v;
    --u, --v;
    g[u].push_back(v), g[v].push_back(u);
  }
  assert(dfs(0));
  sort(ans.begin(), ans.end());
  for (auto P : ans) cout << P.first + 1 << ' ' << P.second + 1 << '\n';
}


### Prompt

Develop a solution in CPP to the problem described below:

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;
typedef long long LL;


const int MAXN = 1e5 + 10;
vector<int> g[MAXN];
bool vis[MAXN] = {0};
bool inst[MAXN] = {0};
vector<pair<int, int>> ans;

bool dfs(int u, int p = -1) {
  vis[u] = 1;
  inst[u] = 1;
  int outdeg = 0;
  for (int v : g[u]) if (v != p) {
    if (not vis[v]) {
      if (dfs(v, u)) ++outdeg, ans.emplace_back(u, v);
    }
    else if (inst[v]) ++outdeg, ans.emplace_back(u, v);
  }
  if (outdeg&1) ans.emplace_back(u, p);
  inst[u] = 0;
  return outdeg % 2 == 0;
}

int main() {
  ios_base::sync_with_stdio(false); cin.tie(0);
  int n, m; cin >> n >> m;
  if (m % 2) cout << -1 << '\n', exit(0);
  for (int i = 0; i < m; ++i) {
    int u, v; cin >> u >> v;
    --u, --v;
    g[u].push_back(v), g[v].push_back(u);
  }
  assert(dfs(0));
  sort(ans.begin(), ans.end());
  for (auto P : ans) cout << P.first + 1 << ' ' << P.second + 1 << '\n';
}


```