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
sys.setrecursionlimit(10**9)

N, M = map(int, input().split())
adjL = [set() for _ in range(N)]
for _ in range(M):
    A, B = map(int, input().split())
    A, B = A-1, B-1
    adjL[A].add(B)

def getCycle(adjL):
    def dfs(vNow):
        if isAvails[vNow]:
            anss.append(vNow)
            return vNow
        useds[vNow] = True
        isAvails[vNow] = True
        for v2 in adjL[vNow]:
            vRet = dfs(v2)
            if vRet == vNow or vRet == numV:
                return numV
            elif vRet != -1:
                anss.append(vNow)
                return vRet
        isAvails[vNow] = False
        return -1

    numV = len(adjL)
    useds = [False] * numV
    anss = []
    for vSt in range(numV):
        isAvails = [False] * numV
        if useds[vSt]: continue
        vRet = dfs(vSt)
        if vRet == numV:
            return anss[::-1]
    return []

cycle = getCycle(adjL)

if not cycle:
    print(-1)
    sys.exit()

isAvails = [False] * N
v2s = [-1] * N
for i in range(len(cycle)):
    isAvails[cycle[i]] = True
    v2s[cycle[i-1]] = cycle[i]

while True:
    adjL2 = [set() for _ in range(N)]
    a, b = -1, -1
    for v in cycle:
        for v2 in adjL[v]:
            if isAvails[v2]:
                adjL2[v].add(v2)
                if v2s[v] != v2:
                    a, b = v, v2

    if (a, b) == (-1, -1):
        print(len(cycle))
        print('\n'.join(map(lambda x: str(x+1), cycle)))
        break

    adjL = adjL2

    cycle = []
    isAvails = [False] * N
    v2s[a] = b
    v = a
    while not isAvails[v]:
        cycle.append(v)
        isAvails[v] = True
        v = v2s[v]