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

stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  # ignore trailing spaces

n, m = na()
g = [[False] * n for _ in range(n)]
gs = [[] for _ in range(n)]
for i in range(m):
    s, t = na()
    gs[s-1].append(t-1)
    g[s-1][t-1] = True

ans = 999999
best = None
for i in range(n):
    ds = [9999999] * n
    prevs = [-1] * n
    ds[i] = 0
    q = [i]
    qp = 0
    while qp < len(q):
        cur = q[qp]
        qp += 1
        for e in gs[cur]:
            if ds[e] > ds[cur] + 1:
                ds[e] = ds[cur] + 1
                prevs[e] = cur
                q.append(e)
    for j in range(n):
        if g[j][i]:
            if ds[j] + 1 < ans:
                ans = ds[j] + 1
                best = [0] * (ds[j]+1)
                best[0] = i
                cur = j
                r = -1
                while cur != i:
                    best[r] = cur
                    r -= 1
                    cur = prevs[cur]

if not best:
    print(-1)
else:
    print(ans)
    for x in best:
        print(x+1)