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 JavaScript (node.js v5.12) (judge verdict: Accepted)

// Atcoder057 - C
function divisor(n) {var ret=[]; for (var i=1;i*i<=n;i++){if(n%i==0){ret.push(i);if(i*i!=n)ret.push(n/i);}}return ret;}
function f(a , b){return Math.max(`${a}`.length, `${b}`.length);}
function main(input) {
  var inputs = input.split('\n');
  var n = parseInt(inputs[0], 10);

  var divisors = divisor(n).sort((a, b) => a - b);
  var surplus = divisors.length % 2;

  var lastIndex = divisors.length - 1;
  var halfLen = Math.floor(divisors.length / 2);

  var minF = `${n}`.length;
  for (var i = 0; i < halfLen - surplus; i++) {
    minF = Math.min(minF, f(divisors[i], divisors[lastIndex - i]));
  }
  if (surplus == 1) minF = Math.min(minF, `${divisors[halfLen]}`.length);

  console.log(minF);
}

main(require('fs').readFileSync('/dev/stdin', 'utf8'));