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:
from collections import deque

N, M = map(int, input().split())

edges = [[] for _ in range(N)]

for _ in range(M):
    fr, to = map(int, input().split())
    fr -= 1
    to -= 1
    edges[fr].append(to)

for start in range(N):
    st = deque([(start, True)])
    L = deque()
    V = set()

    while st:
        now, isFirst = st.pop()
        if isFirst:
            L.append(now)

            if now in V:
                break

            V.add(now)

            st.append((now, False))

            for to in edges[now]:
                st.append((to, True))
        else:
            L.pop()
            V.remove(now)
    else:
        continue

    loopStart = L.pop()
    while L[0] != loopStart:
        L.popleft()

    ans = []
    K = len(L)
    for i, now in enumerate(L):
        for to in edges[now]:
            if i + 1 >= K or to == L[i + 1]:
                ans.append(now + 1)
                break
        else:
            break
    else:
        print(0)
        print()
        # print(len(ans))
        # print(*ans, sep='\n')
        exit()

print(-1)