You are given a positive integer n, written without leading zeroes (for example, the number 04 is incorrect). 

In one operation you can delete any digit of the given integer so that the result remains a positive integer without leading zeros.

Determine the minimum number of operations that you need to consistently apply to the given integer n to make from it the square of some positive integer or report that it is impossible.

An integer x is the square of some positive integer if and only if x=y^2 for some positive integer y.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{9}). The number is given without leading zeroes.

Output

If it is impossible to make the square of some positive integer from n, print -1. In the other case, print the minimal number of operations required to do it.

Examples

Input

8314


Output

2


Input

625


Output

0


Input

333


Output

-1

Note

In the first example we should delete from 8314 the digits 3 and 4. After that 8314 become equals to 81, which is the square of the integer 9.

In the second example the given 625 is the square of the integer 25, so you should not delete anything. 

In the third example it is impossible to make the square from 333, so the answer is -1.
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###################################The_Apurv_Rathore#####################################################
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import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math 


alphabets = list('abcdefghijklmnopqrstuvwxyz')


def isPrime(x):
    for i in range(2,x):
        if i*i>x:
            break
        if (x%i==0):
            return False
    return True
def ncr(n, r, p):  
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,
                      p - 2, p)) % p
def primeFactors(n): 
    l = []
    while n % 2 == 0: 
        l.append(2)
        n = n / 2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            l.append(int(i))
            n = n / i 
    if n > 2: 
        l.append(n)
    return list(set(l))
def power(x, y, p) : 
	res = 1
	x = x % p 
	if (x == 0) : 
		return 0
	while (y > 0) : 
		if ((y & 1) == 1) : 
			res = (res * x) % p 
		y = y >> 1	 # y = y/2 
		x = (x * x) % p 		
	return res 
def SieveOfEratosthenes(n):
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime
def countdig(n):
    c = 0
    while (n > 0):
        n //= 10
        c += 1
    return c
def si():
    return input()
def prefix_sum(arr):
    r = [0] * (len(arr)+1)
    for i, el in enumerate(arr):
        r[i+1] = r[i] + el
    return r
def divideCeil(n,x):
    if (n%x==0):
        return n//x
    return n//x+1
def ii():
    return int(input())
def li():
    return list(map(int,input().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def fastPrintNextLines(a):
    # 12
    # 3
    # 1
    #like this
    #a is list of strings
    print('\n'.join(map(str,a)))

#__________________________TEMPLATE__________________OVER_______________________________________________________

from itertools import chain, combinations
def power_set(L):
    """
    L is a list.
    The function returns the power set, but as a list of lists.
    """
    cardinality=len(L)
    n=2 ** cardinality
    powerset = []
    
    for i in range(n):
        a=bin(i)[2:]
        subset=[]
        for j in range(len(a)):
            if a[-j-1]=='1':
                subset.append(L[j])
        powerset.append(subset)
        
    #the function could stop here closing with
    #return powerset

    powerset_orderred=[]
    for k in range(cardinality+1):
        for w in powerset:
            if len(w)==k:
                powerset_orderred.append(w)
        
    return powerset_orderred

if(os.path.exists('input.txt')):
    sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
# else:
#     input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = 1
for i in range(t):
    s = list(si())
    q = len(s)
    s = list(power_set(s))
    m = 1000000000000000000000
    for i in range(len(s)):
        n = ''.join(s[i])
        if (n==''):
            continue
        
        n = int(n)
        if (len(str(n))!=len(s[i])) or n==0:
            # print(n)
            continue
        x = n**0.5

        if x==int(x):
            # print(n)
            # print(s[i])
            # print(q,len(s[i]))
            m = min(m,q - len(s[i]))
    if (m==1000000000000000000000):
        m = -1
    print(m)