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: Wrong Answer)


var lines = [];
var readline = require('readline');
var C = [].fill(0);
var result = Infinity;

var rl = readline.createInterface({
    input: process.stdin,
    output: process.stdout,
});

rl.on('line', function (x) {
    lines.push(x);
})

rl.on('close', function () {
    var N = Number(lines.shift()); var n = N;
    var counter = [].fill(0);

    for (var i = 2; i < Math.sqrt(N); i++) {
        if (counter[i] === undefined && n % i === 0) { counter[i] = 0; }
        while (n % i === 0) {
            n = n / i;
            counter[i]++;
        }
    }

    if (counter.length === 0) console.log(N.toString().length);
    else {
        
        counter.forEach(function (value, index, array) {
            for (var i = 0; i < value; i++) {
                C.push(index);
            }
        });
        if (n !== 1) C.push(n);
        dfs(0, []);
        console.log(result);
    }
});

function dfs(n, comb) {
    if (n === C.length) { calc(comb); return (0); }

    comb[n] = 0;
    dfs(n + 1, comb);

    comb[n] = 1;
    dfs(n + 1, comb);
}

function calc(comb) {
    var a = 1; var b = 1;

    for (var i = 0; i < comb.length; i++) {
        if (comb[i] === 0) { a *= C[i]; }
        else { b *= C[i]; }
    }
    if (result > b.toString().length && a.toString().length <= b.toString().length) { result = b.toString().length; }
}