#include <iostream>
#include <vector>

using namespace std;

bool printed = false;
vector<vector<int>> graph;
vector<vector<bool>> edges;

void dfs(int v, vector<int> path, vector<bool> used) {
    for (int i = path.size() - 1; !printed && i >= 0; i--) {
        if (edges[v][path[i]]) {
            path.push_back(v);
            cout << path.size() - i << endl;
            for (int j = i; j < path.size(); j++)cout << path[j] + 1 << endl;
            printed = true;
            return;
        }
    }
    if (printed)return;

    vector<bool> next_used = used;
    for (auto x:graph[v])next_used[x] = true;

    for (auto nextv:graph[v]) {
        if (!used[nextv]) {
            path.push_back(v);
            dfs(nextv, path, next_used);
            path.pop_back();
        }
    }
}

int main() {
    int n, m;
    cin >> n >> m;

    graph = vector<vector<int>>(n);
    edges = vector<vector<bool>>(n, vector<bool>(n));
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        graph[a].push_back(b);
        edges[a][b] = true;
    }

    for (int i = 0; i < n; i++)dfs(i, {}, vector<bool>(n, false));
    if (!printed)cout << -1 << endl;
    return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

Given is a directed graph G with N vertices and M edges.
The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i.
It is guaranteed that the graph contains no self-loops or multiple edges.

Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph.
Here the null graph is not considered as a subgraph.

Constraints

* 1 \leq N \leq 1000
* 0 \leq M \leq 2000
* 1 \leq A_i,B_i \leq N
* A_i \neq B_i
* All pairs (A_i, B_i) are distinct.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M
A_1 B_1
A_2 B_2
:
A_M B_M


Output

If there is no induced subgraph of G that satisfies the condition, print `-1`. Otherwise, print an induced subgraph of G that satisfies the condition, in the following format:


K
v_1
v_2
:
v_K


This represents the induced subgraph of G with K vertices whose vertex set is \\{v_1, v_2, \ldots, v_K\\}. (The order of v_1, v_2, \ldots, v_K does not matter.) If there are multiple subgraphs of G that satisfy the condition, printing any of them is accepted.

Examples

Input

4 5
1 2
2 3
2 4
4 1
4 3


Output

3
1
2
4


Input

4 5
1 2
2 3
2 4
1 4
4 3


Output

-1


Input

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2


Output

4
2
3
4
5

### Response

```cpp
#include <iostream>
#include <vector>

using namespace std;

bool printed = false;
vector<vector<int>> graph;
vector<vector<bool>> edges;

void dfs(int v, vector<int> path, vector<bool> used) {
    for (int i = path.size() - 1; !printed && i >= 0; i--) {
        if (edges[v][path[i]]) {
            path.push_back(v);
            cout << path.size() - i << endl;
            for (int j = i; j < path.size(); j++)cout << path[j] + 1 << endl;
            printed = true;
            return;
        }
    }
    if (printed)return;

    vector<bool> next_used = used;
    for (auto x:graph[v])next_used[x] = true;

    for (auto nextv:graph[v]) {
        if (!used[nextv]) {
            path.push_back(v);
            dfs(nextv, path, next_used);
            path.pop_back();
        }
    }
}

int main() {
    int n, m;
    cin >> n >> m;

    graph = vector<vector<int>>(n);
    edges = vector<vector<bool>>(n, vector<bool>(n));
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        graph[a].push_back(b);
        edges[a][b] = true;
    }

    for (int i = 0; i < n; i++)dfs(i, {}, vector<bool>(n, false));
    if (!printed)cout << -1 << endl;
    return 0;
}
```