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

const int nax = 1e5;
int bad[nax];
struct Edge {
  int s, e;
};
vector<Edge> edges;
vector<int> node[nax];

int done[nax];
vector<pair<int,int>> topo;
void dfs(int p, int par_ei = -1) {
  if (done[p]++) return;
  for (int ei : node[p]) {
    int i = edges[ei].e;
    dfs(i, ei);
  }
  topo.push_back({p, par_ei});
}

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, m;
  cin >> n >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    node[a].push_back(edges.size());
    edges.push_back({a,b});
    node[b].push_back(edges.size());
    edges.push_back({b,a});
    bad[a] ^= 1;
  }
  int bads = 0;
  for (int i = 0; i < n; i++)
    bads += bad[i];
  if (bads%2) {
    cout << -1 << endl;
    return 0;
  }
  dfs(0);
  for (auto p : topo) {
    int i = p.first, ei = p.second;
    if (bad[i]) {
      bad[i] ^= 1;
      bad[edges[ei].s] ^= 1;
      swap(edges[ei], edges[ei^1]);
    }
  }
  for (int i = 0; i < edges.size(); i += 2)
    cout << edges[i].s+1 << ' ' << edges[i].e+1 << '\n';
}

### 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;
typedef long long ll;

const int nax = 1e5;
int bad[nax];
struct Edge {
  int s, e;
};
vector<Edge> edges;
vector<int> node[nax];

int done[nax];
vector<pair<int,int>> topo;
void dfs(int p, int par_ei = -1) {
  if (done[p]++) return;
  for (int ei : node[p]) {
    int i = edges[ei].e;
    dfs(i, ei);
  }
  topo.push_back({p, par_ei});
}

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, m;
  cin >> n >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    node[a].push_back(edges.size());
    edges.push_back({a,b});
    node[b].push_back(edges.size());
    edges.push_back({b,a});
    bad[a] ^= 1;
  }
  int bads = 0;
  for (int i = 0; i < n; i++)
    bads += bad[i];
  if (bads%2) {
    cout << -1 << endl;
    return 0;
  }
  dfs(0);
  for (auto p : topo) {
    int i = p.first, ei = p.second;
    if (bad[i]) {
      bad[i] ^= 1;
      bad[edges[ei].s] ^= 1;
      swap(edges[ei], edges[ei^1]);
    }
  }
  for (int i = 0; i < edges.size(); i += 2)
    cout << edges[i].s+1 << ' ' << edges[i].e+1 << '\n';
}

```