Problem:
Score : 300 points

Problem Statement

You are given an integer N.

For two positive integers A and B, we will define F(A,B) as the larger of the following: the number of digits in the decimal notation of A, and the number of digits in the decimal notation of B.

For example, F(3,11) = 2 since 3 has one digit and 11 has two digits.

Find the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Constraints

1 \leq N \leq 10^{10}

N is an integer.

Input

The input is given from Standard Input in the following format:

N

Output

Print the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Sample Input 1

10000

Sample Output 1

3

F(A,B) has a minimum value of 3 at (A,B)=(100,100).

Sample Input 2

1000003

Sample Output 2

7

There are two pairs (A,B) that satisfy the condition: (1,1000003) and (1000003,1). For these pairs, F(1,1000003)=F(1000003,1)=7.

Sample Input 3

9876543210

Sample Output 3

6

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

n = int(input())
import math
sq = math.sqrt(n)
if isinstance(sq, int):
    print(len(str(sq)))
    quit()
"""nを素因数分解"""
"""2以上の整数n => [[素因数, 指数], ...]の2次元リスト"""
# n == 1のとき、[1, 1]を返す
def factorization(n):
    
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            #cnt=0
            while temp%i==0:
                #cnt+=1
                temp //= i
                arr.append(i)
            #arr.append([i, cnt])

    if temp!=1:
        #arr.append([temp, 1])
        arr.append(temp)

    if arr==[]:
        #arr.append([n, 1])
        arr.append(n)
    return arr
factor = factorization(n)
if len(factor) == 1:
    print(len(str(n)))
    quit()
first = 1
ans = float('Inf')
for i in range(len(factor) - 1):
    first *= factor[i]
    second = int(n / first)
    #print(first, second)
    ans = min(ans, max(len(str(first)), len(str(second))))
print(ans)