51_D. Geometrical problem
Polycarp loves geometric progressions — he collects them. However, as such progressions occur very rarely, he also loves the sequences of numbers where it is enough to delete a single element to get a geometric progression.

In this task we shall define geometric progressions as finite sequences of numbers a1, a2, ..., ak, where ai = c·bi - 1 for some real numbers c and b. For example, the sequences [2, -4, 8], [0, 0, 0, 0], [199] are geometric progressions and [0, 1, 2, 3] is not.

Recently Polycarp has found a sequence and he can't classify it. Help him to do it. Determine whether it is a geometric progression. If it is not, check if it can become a geometric progression if an element is deleted from it.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of elements in the given sequence. The second line contains the given sequence. The numbers are space-separated. All the elements of the given sequence are integers and their absolute value does not exceed 104.

Output

Print 0, if the given sequence is a geometric progression. Otherwise, check if it is possible to make the sequence a geometric progression by deleting a single element. If it is possible, print 1. If it is impossible, print 2.

Examples

Input

4
3 6 12 24


Output

0


Input

4
-8 -16 24 -32


Output

1


Input

4
0 1 2 3


Output

2
```python
def main():
    n = int(input())
    l = tuple(map(int, input().split()))
    if n == 1:
        return 0
    if n == 2:
        if l[0] == 0 and l[1] != 0: return 1
        return 0
    def div(a, b):
        if b == 0: return 0 if a == 0 else "inf"
        return a / b
    pref, suff = [["init", 1]], [["init", 1]]
    for i in range(1, n):
        d = div(l[i], l[i-1])
        if pref[-1][0] == "init":
            pass
        elif d != pref[-1][0]:
            for j in range(i, n):
                pref += [["null", 0]]
            break
        pref += [[d, 1]]
    for i in range(n - 1, -1, -1):
        d = div(l[i], l[i-1])
        if l[i - 1] == 0 and l[i] != 0:
            suff[-1][0] = suff[-2][0] if len(suff) > 1 else 1
        if d != suff[-1][0] and suff[-1][0] != "init":
            for j in range(i-1, -1,-1):
                suff += [["null", 0]]
            break
        suff += [[d, 1]]
    #print(pref)
    #print(suff)
    if pref[-1][1] == 1: return 0
    if pref[-2][1] == 1 or suff[-2][1] == 1: return 1
    
    for i in range(1,n-1):
        pr, sf = pref[i - 1], suff[n - i - 2]
        dv = div(l[i+1], l[i-1])
        #print(pr,sf,dv,l[i-1:i+2])
        if pr[0] == "init" and sf[0] == "init": return 1
        if pr[0] == "init" and dv == sf[0]:
            return 1
        elif sf[0] == "init" and dv == pr[0]:
            return 1
        elif dv == pr[0] and dv == sf[0]:
            return 1
    return 2
print(main())

```