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
from collections import deque


def solve():
    N, M = map(int, input().split())
    edges = [set() for _ in range(N+1)]
    for a, b in (map(int, line.split()) for line in sys.stdin):
        edges[a].add(b)

    # 答えが見つかるまで、スタート地点は全て試します
    for start in range(1, N+1):
        visited = [0]*(N+1)

        dq = deque([(start, {start})])
        visited[start] = 1

        while dq:
            # 点と、これまで通ってきた頂点を取り出す
            v, route = dq.popleft()
            # これまで通ってきた頂点集合と、行き先の頂点集合の積集合をとる
            s = route & edges[v]
            # s=={start}でなければuとおr sが空集合でも通る。sが何やらいろいろあると、通れない。
            if not (not s or s == {start}):
                continue
            for dest in edges[v]:
                # startに戻ってきたら出力して終了
                if dest == start:
                    print(len(route))
                    print(*route, sep='\n')
                    exit()
                # 訪問済みであれば、スルー
                if visited[dest]:
                    continue
                visited[dest] = 1
                # destを始点として、通ってきた頂点にdestを追加したもの
                dq.append((dest, route | {dest}))

    print(-1)


if __name__ == '__main__':
    solve()