Problem:
Score : 300 points

Problem Statement

Takahashi is standing on a multiplication table with infinitely many rows and columns.

The square (i,j) contains the integer i \times j. Initially, Takahashi is standing at (1,1).

In one move, he can move from (i,j) to either (i+1,j) or (i,j+1).

Given an integer N, find the minimum number of moves needed to reach a square that contains N.

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum number of moves needed to reach a square that contains the integer N.

Sample Input 1

10

Sample Output 1

5

(2,5) can be reached in five moves. We cannot reach a square that contains 10 in less than five moves.

Sample Input 2

50

Sample Output 2

13

(5, 10) can be reached in 13 moves.

Sample Input 3

10000000019

Sample Output 3

10000000018

Both input and output may be enormous.

Solution in JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

var GET = (function () {function f(s) {return new g(s);}function g(s) {this._s = s.trim().split("\n");}g.prototype.getValue = function (n) {return this._s[n];};g.prototype.getArray = function (n) {return this.stoa(this._s[n]);};g.prototype.slice = function (s, limit) {return this._s.slice(s, +limit + s || undefined).map(v => this.stox(v));};g.prototype.sliceArray = function (s, limit) {return this._s.slice(s, +limit + s || undefined).map(v => this.stoa(v));};g.prototype.stoa = function (s) {return s.split(" ").map(v => this.stox(v));};g.prototype.stox = function (v) {var i = +v;return (i === i) ? i : v;};return f;})();
var o = GET(require("fs").readFileSync("/dev/stdin", "utf8"));
function PriorityQueue(){this.heap = [];}PriorityQueue.prototype = {push : function(x){var h = this.heap, i = h.length++, j;while(i){j = i-1 >> 1;if(h[j] <= x)break;h[i] = h[j];i = j;}h[i] = x;}, pop : function(){var h = this.heap, r = h[0], x = h.pop();var i = 0, k = h.length >> 1, j;while (i < k) {j = (i << 1) + 1;if(h[j+1] < h[j])++j;if(x <= h[j])break;h[i] = h[j];i = j;}if(h.length)h[i] = x;return r;}, size : function(){return this.heap.length;}, top : function(){return this.heap[0];},};

console.log(main());
function main(){
    var N = o.getValue(0);

    var ans = 0;
    for (var i = 1; i * i < N; i++) {
        if (N % i !== 0) continue;
        var j = N / i;
        if (ans === 0) {
            ans = i+j-2;
        } else {
            ans = Math.min(ans, i + j - 2);
        }
    }

    return ans;
}