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 heapq


class Edge():

    def __init__(self, _to, _cost):
        self.to = _to
        self.cost = _cost


class Dijkstra():
    def __init__(self, V):

        self.G = [[] for i in range(V+1)]
        self._E = 0
        self._V = V

    @property
    def E(self):
        """ 辺数 """
        return self._E

    @property
    def V(self):
        """ 頂点数 """
        return self._V

    def add(self, _from, _to, _cost):
        self.G[_from].append(Edge(_to, _cost))
        self._E += 1

    def shortest_path(self, s):

        que = []
        heapq.heapify(que)
        d = [10**10] * (self._V+1)
        par = [-1] * (self._V + 1)
        d[s] = 0
        heapq.heappush(que, (0, s))

        while que:
            cost, v = heapq.heappop(que)
            if d[v] < cost:
                continue

            for e in self.G[v]:
                if d[e.to] > d[v] + e.cost:
                    d[e.to] = d[v] + e.cost
                    par[e.to] = v
                    heapq.heappush(que, (d[e.to], e.to))
        return d, par


N, M = map(int, input().split())
G = []
ans = []
now_dis = 10**6
for i in range(M):
    a, b = map(int, input().split())
    G.append((a, b))
for i in range(M):
    graph = Dijkstra(N)
    for j in range(M):
        if j != i:
            graph.add(G[j][0], G[j][1], 1)
    S, T = G[i][0], G[i][1]
    d, par = graph.shortest_path(T)
    if d[S]+1 < now_dis:
        now_dis = d[S]+1
        sub = []
        node = S
        while node != T:
            sub.append(node)
            node = par[node]
        sub.append(T)
        ans = sub

K=len(ans)
if K==0:
    print(-1)
    exit()
print(K)
for x in ans:
    print(x)