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];

// visited edges
static char visited[N];

struct entry {
    int prev;
    int start;
    int end;
};

static int map[N][N];

static struct entry e_heap[M];
static int e_cur;
static int e_ind;

struct answer {
    struct entry ent;
    int size;
};

struct answer solve() {
    struct answer ans;
    struct entry ent, base, cur, min_cur;
    int min = N+1, count;

    while (e_cur < e_ind) {
        cur = e_heap[e_cur++];

        for (int i = 0; i < n; i++) {
            if (edges[i][cur.end] == 1) {
                // printf("%d -> %d\n", cur.end, i);
                ent.prev = cur.start;
                ent.start = cur.end;
                ent.end = i;
                map[ent.end][ent.start] = ent.prev;
                e_heap[e_ind++] = ent;
                visited[ent.start] = 1;

                base = ent;

                count = 0;
                while (ent.start != ent.prev) {
                    int prev = ent.prev;
                    ent.end = ent.start;
                    ent.start = prev;
                    ent.prev = map[ent.end][ent.start];
                    count++;
                    // printf("  %d -> %d\n", ent.start, ent.end);
                    if (base.end == ent.start) {
                        // printf("count=%d, prev=%d, start=%d, end=%d\n",
                        //         count, ent.prev, ent.start, ent.end);
                        if (count + 1 < min) {
                            min = count + 1;
                            min_cur = base;
                        }
                        break;
                    }
                }
            }
        }

        if (min < N) {
            ans.ent = min_cur;
            ans.size = min;
            return ans;
        }
    }

    ans.size = -1;
    return ans;
}

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;
    }

    struct entry ent;
    struct answer ans, cand;
    ans.size = -1;

    for (;;) {
        cand = solve();
        if (cand.size != -1 && (ans.size < 0 || cand.size < ans.size)) {
            ans = cand;
        }

        int all_visited = 1;
        for (int v = 0; v < n; v++) {
            if (visited[v] == 0) {
                all_visited = 0;
                visited[v] = 1;
                for (int i = 0; i < n; i++) {
                    if (edges[i][v] == 1) {
                        ent.prev = v;
                        ent.start = v;
                        ent.end = i;
                        e_heap[e_ind++] = ent;
                    }
                }
            }
        }
        if (all_visited) {
            break;
        }
    }

    printf("%d\n", ans.size);
    if (ans.size >= 1) {
        ent = ans.ent;

        for (int i = 0; i < ans.size; i++) {
            int prev = ent.prev;

            // printf("prev=%d, start=%d, end=%d\n", ent.prev, ent.start, ent.end);
            printf("%d\n", ent.start+1);
            ent.end = ent.start;
            ent.start = prev;
            ent.prev = map[ent.end][ent.start];
        }
    }

    return 0;
}