Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, …, x_{k-1}.

Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 ≤ i ≤ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.

For example, if the x = [1, 2, 3] and n = 5, then:

  * a_0 = 0, 
  * a_1 = x_{0mod 3}+a_0=x_0+0=1, 
  * a_2 = x_{1mod 3}+a_1=x_1+1=3, 
  * a_3 = x_{2mod 3}+a_2=x_2+3=6, 
  * a_4 = x_{3mod 3}+a_3=x_0+6=7, 
  * a_5 = x_{4mod 3}+a_4=x_1+7=9. 



So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].

Now the boy hopes that he will be able to restore x from a! Knowing that 1 ≤ k ≤ n, help him and find all possible values of k — possible lengths of the lost array.

Input

The first line contains exactly one integer n (1 ≤ n ≤ 1000) — the length of the array a, excluding the element a_0.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).

Note that a_0 is always 0 and is not given in the input.

Output

The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.

The second line of the output should contain l integers — possible lengths of the lost array in increasing order.

Examples

Input

5
1 2 3 4 5


Output

5
1 2 3 4 5 

Input

5
1 3 5 6 8


Output

2
3 5 

Input

3
1 5 3


Output

1
3 

Note

In the first example, any k is suitable, since a is an arithmetic progression.

Possible arrays x:

  * [1]
  * [1, 1]
  * [1, 1, 1]
  * [1, 1, 1, 1] 
  * [1, 1, 1, 1, 1]



In the second example, Bajtek's array can have three or five elements.

Possible arrays x:

  * [1, 2, 2]
  * [1, 2, 2, 1, 2]



For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.

In the third example, only k = n is good.

Array [1, 4, -2] satisfies the requirements.

Note that x_i may be negative.
import math
import bisect
import itertools
import sys
'''mod=10**9 +7
fact=[1]*1001
ifact=[1]*1001
for i in range(1,1001):
    fact[i]=((fact[i-1])*i)%mod
    ifact[i]=((ifact[i-1])*pow(i,mod-2,mod))%mod
def ncr(n,r):
    return (((fact[n]*ifact[n-r])%mod)*ifact[r])%mod
def npr(n,r):
    return (((fact[n]*ifact[n-r])%mod))
    '''


def mindiff(a):
    b=a[:]
    b.sort()
    m=10000000000
    for i in range(len(b)-1):
        if b[i+1]-b[i]<m:
            m=b[i+1]-b[i]
    return m
    
def lcm(a,b):
    return a*b//math.gcd(a,b)

    
def merge(a,b):
    i=0;j=0
    c=0
    ans=[]
    while i<len(a) and j<len(b):
        if a[i]<b[j]:
            ans.append(a[i])
            i+=1
        else:
            ans.append(b[j])
            c+=len(a)-i
            j+=1
    ans+=a[i:]
    ans+=b[j:]
    return ans,c
def mergesort(a):
    if len(a)==1:
        return a,0
    mid=len(a)//2   
    left,left_inversion=mergesort(a[:mid])
    right,right_inversion=mergesort(a[mid:])
    m,c=merge(left,right)
    c+=(left_inversion+right_inversion)
    return m,c
    
def is_prime(num):
    if num == 2: return True
    if num == 3: return True
    if num%2 == 0: return False
    if num%3 == 0: return False
    t = 5
    a = 2
    while t <= int(math.sqrt(num)):
        if num%t == 0: return False
        t += a
        a = 6 - a
    return True
    
  
def ceil(a,b):
    if a%b==0:
        return a//b
    else:
        return (a//b + 1)

def binsearch(arr,b,low,high):
    if low==high:
        return low
    if arr[math.ceil((low+high)/2)]<b:
        return binsearch(arr,b,low,math.ceil((low+high)/2) -1 )
    else:
        return binsearch(arr,b,math.ceil((low+high)/2),high)
    

#/////////////////////////////////////////////////////////////////////////////////////////////////


n=int(input())
a=[];x=[]
a.append(0)
a+=list(map(int,input().split()))
for i in range(1,n+1):
    x.append(a[i]-a[i-1])
c=[];
#print(x)
for i in range(n):
    b=x[:i+1]
    if x[:n+1]==b*(n//(i+1)) + b[:n%(i+1)]:
        c.append(i+1)
    #print(b)    
print(len(c))
print(*c)
    