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

Solution (in PYTHON3):

Solution:
import sys

read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines
sys.setrecursionlimit(10 ** 9)
INF = 1 << 60
MOD = 1000000007


def main():
    N, M, *AB = map(int, read().split())
    G = [[] for _ in range(N)]
    for a, b in zip(*[iter(AB)] * 2):
        G[a - 1].append(b - 1)

    def dfs(v, p, seen, finished, hist):
        seen[v] = True
        hist.append(v)
        for nv in G[v]:
            if nv == p:
                continue
            if finished[nv]:
                continue
            if seen[nv] and not finished[nv]:
                return nv

            pos = dfs(nv, v, seen, finished, hist)
            if pos != -1:
                return pos
        hist.pop()
        finished[v] = True
        return -1

    for i in range(N):
        seen = [False] * N
        finished = [False] * N
        hist = []
        pos = dfs(i, -1, seen, finished, hist)
        if pos != -1:
            break

    if pos == -1:
        print(-1)
        return

    cycle = []
    while hist:
        t = hist.pop()
        cycle.append(t)
        if t == pos:
            break

    cycle.reverse()

    while True:
        order = [-1] * N
        for i, v in enumerate(cycle):
            order[v] = i

        for i, v in enumerate(cycle):
            ord_from = ord_to = -1
            for nv in G[v]:
                if v == nv or order[nv] == -1 or order[nv] == (order[v] + 1) % len(cycle):
                    continue
                ord_from, ord_to = i, order[nv]
                break
            if ord_from != -1:
                break

        if ord_from == -1:
            break

        if ord_from < ord_to:
            cycle = cycle[: ord_from + 1] + cycle[ord_to:]
        else:
            cycle = cycle[ord_to : ord_from + 1]

    print(len(cycle))
    for v in cycle:
        print(v + 1)

    return


if __name__ == '__main__':
    main()