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

     
def dfs(N, AB):
    status = [-1] * N
    for i in range(N):
        if status[i] == 1:
            continue
        stack = [i]
        status[i] = 0
        while stack:
            v = stack[-1]
            if AB[v]:
                n = AB[v].popleft()
                if status[n] == -1:
                    stack.append(n)
                    status[n] = 0
                else:
                    if n in stack:
                        idx = stack.index(n)
                        cycle = stack[idx:]
                        return cycle
            else:                  
                status[v] = 1
                stack.pop()
    return False


def find_smaller_cycle(cycle, AB):
    i = 0
    while i < len(cycle):
        v = cycle[i]
        if AB[v]:
            n = AB[v].popleft()
            if n in cycle:
                r = cycle.index(n)
                if i < r:
                    cycle = cycle[:i+1] + cycle[r:]
                else:
                    cycle = cycle[r:i+1]
                i = cycle.index(v)
        else:
            i += 1
            
    return cycle


N, M = [int(i) for i in input().split()]
AB = [deque() for _ in range(N)]
for _ in range(M):
    A, B = [int(i) - 1 for i in input().split()]
    AB[A].append(B)
                      
cycle = dfs(N, AB)
if not cycle:
    print(-1)
else:
    cycle = find_smaller_cycle(cycle, AB)
            
    print(len(cycle))
    for v in cycle:
        print(v + 1)