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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

// Max verticles
#define N 1000
// Max edges
#define M 2000

// Num of verticles
static int n;
// Num of edges
static int m;

// edges[1][0] == 1 means theare is a edge from v1 to v2.
static char edges[N][N];

#define push(stack, ind, elem) (stack[ind++] = elem)
#define pop(stack, ind) (ind--)

static char v_stack[N];
static int v_ind;

// static char e_stack[M];
// static int e_ind;

int solve(int start) {
    for (int i = 0; i < v_ind; i++) {
        if (v_stack[i] == start) {
            return 1;
        }
    }

    push(v_stack, v_ind, start);

    for (int i = 0; i < n; i++) {
        if (edges[i][start] == 1) {
            if (solve(i)) {
                return 1;
            }
        }
    }

    pop(v_stack, v_ind);

    return 0;
}

int main(void) {
    scanf("%d %d", &n, &m);

    for (int i = 0; i < m; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        edges[b-1][a-1] = 1;
    }

    if (solve(0)) {
        printf("%d\n", v_ind);
        for (int i = 0; i < v_ind; i++) {
            printf("%d\n", v_stack[i] + 1);
        }
    } else {
        printf("%d\n", -1);
    }

    return 0;
}