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



Sample input:
4 5
1 2
2 3
2 4
4 1
4 3


Sample output:
3
1
2
4

Submission status: Runtime Error

Solution code:
def dfs(node, graph, visited, stack):
    visited[node] = True
    stack.append(node)
    if node in graph:
        for n in graph[node]:
            if n not in stack:
                if not visited[n]:
                    dfs(n, graph, visited, stack)
            else:
                index = stack.index(n)
                print(len(stack) - index)
                for i in stack[index:]:
                    print(i)
                #print(n)
                exit(0)

    stack.pop(-1)


def main():
    graph = {}
    visited = {}
    stack = []
    emp, num = (int(i) for i in input().split())  
    for i in range(num):
        n1, n2 = (int(i) for i in input().split())  
        if n1 not in graph:
            graph[n1] = [n2]
        elif n2 not in graph[n1]:
            graph[n1].append(n2)
        if n1 not in visited:
            visited[n1] = False
        if n2 not in visited:
            visited[n2] = False

    for node in list(visited.keys()):
        if not visited[node]:
            dfs(node, graph, visited, stack)

if __name__ == "__main__":
    main()
    print(-1)

Submission status: Runtime Error

Solution code:
import sys

def dfs(node, graph, visited, stack):
    visited[node] = True
    stack.append(node)
    if node in graph:
        for n in graph[node]:
            if n not in stack:
                if not visited[n]:
                    dfs(n, graph, visited, stack)
            else:
                index = stack.index(n)
                print(len(stack) - index)
                for i in stack[index:]:
                    print(i)
                #print(n)
                exit(0)

    stack.pop(-1)


def main():
    graph = {}
    visited = {}
    stack = []
    emp = int(sys.stdin.readline())
    num = int(sys.stdin.readline())
    for i in range(num):
        n1, n2 = sys.stdin.readline().strip().split(' ')
        if n1 not in graph:
            graph[n1] = [n2]
        elif n2 not in graph[n1]:
            graph[n1].append(n2)
        if n1 not in visited:
            visited[n1] = False
        if n2 not in visited:
            visited[n2] = False

    for node in list(visited.keys()):
        if not visited[node]:
            dfs(node, graph, visited, stack)

if __name__ == "__main__":
    main()
    print(-1)