Problem:
Score : 600 points

Problem Statement

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.

Notes

For a directed graph G = (V, E), we call a directed graph G' = (V', E') satisfying the following conditions an induced subgraph of G:

V' is a (non-empty) subset of V.

E' is the set of all the edges in E that have both endpoints in V'.

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.

Sample Input 1

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

Sample Output 1

3
1
2
4

The induced subgraph of G whose vertex set is \{1, 2, 4\} has the edge set \{(1, 2), (2, 4), (4, 1)\}. The in-degree and out-degree of every vertex in this graph are both 1.

Sample Input 2

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

Sample Output 2

-1

There is no induced subgraph of G that satisfies the condition.

Sample Input 3

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

Sample Output 3

4
2
3
4
5

Solution in Rust (1.42.0) (judge verdict: Time Limit Exceeded)

#![allow(unused_imports)]
use {
   proconio::{fastout, input, marker::*},
   std::cmp::*,
   std::collections::*,
};

fn main() {
   input! {
      n: usize,
      m: usize,
      ab: [(Usize1, Usize1); m]
   }

   let mut edge = vec![vec![false; n]; n];
   let mut to = vec![vec![]; n];
   for (a, b) in ab {
      to[a].push(b);
      edge[a][b] = true;
   }
   for i in 0..n {
      let mut nodes = vec![];
      let mut used = vec![false; n];
      if !dfs(i, i, &mut used, &mut nodes, &to) {
         continue;
      }
      nodes.reverse();
      loop {
         let n = nodes.len();
         let mut update = None;
         'outer: for i in 0..n {
            for j in 0..n {
               if i == j || j == (i + 1) % n {
                  continue;
               }
               if edge[nodes[i]][nodes[j]] {
                  update = Some((i, j));
                  break 'outer;
               }
            }
         }
         if let Some((i, j)) = update {
            let mut new = vec![];
            if i < j {
               for k in 0..n {
                  if i < k && k < j {
                     continue;
                  }
                  new.push(nodes[k]);
               }
            } else {
               for k in 0..n {
                  if k < i || j < k {
                     continue;
                  }
                  new.push(nodes[k]);
               }
            }
            nodes = new;
         } else {
            break;
         }
      }
      println!("{}", nodes.len());
      for a in nodes {
         println!("{}", a + 1);
      }
      return;
   }
   println!("-1");
}

fn dfs(
   v: usize,
   t: usize,
   used: &mut Vec<bool>,
   nodes: &mut Vec<usize>,
   to: &Vec<Vec<usize>>,
) -> bool {
   used[v] = true;
   for &u in &to[v] {
      if u == t || !used[u] && dfs(u, t, used, nodes, to) {
         nodes.push(v);
         return true;
      }
   }
   used[v] = false;
   false
}