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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <boost/foreach.hpp>
using namespace std;
#define pi 2*asin(1.0)
#define MAX 10000000
#define MOD 1000000007
#define _GLIBCXX_DEBUG
 
vector<long> fac(MAX);
vector<long> finv(MAX);
vector<long> inv(MAX);
vector<long> Primelist;
//template functions
/////////////////////////
void Prinit(){
long PR=10000001;
vector<long> P(PR,0);
for(long i=2;i*i<PR;i++){
    if(P.at(i)==0){
        long j=i;
      j+=i;
        while(j<PR){
            P.at(j)=1;
             j+=i;         
        }
    }
}

for(long i=2;i*i<PR;i++){//Primelist.size()=446
    if(P.at(i)==0){
        Primelist.push_back(i);
    }
}
}

void Cominit(){
  fac.at(0)=fac.at(1)=1;
  finv.at(0)=finv.at(1)=1;
  inv.at(1)=1;
  for(long i=2;i<MAX;i++){
      fac.at(i)=(fac.at(i-1)*i)%MOD;
      inv.at(i)=MOD-inv.at(MOD%i)*(MOD/i)%MOD;
      finv.at(i)=finv.at(i-1)*inv.at(i)%MOD;
  }
}

long facComb(long N,long M){// nCm%
if(N<M) return 0;
if(N<0||M<0) return 0;
return fac.at(N)*(finv.at(M)*finv.at(N-M)%MOD)%MOD;
}

long Comb(long N,long M){// nCm
long ans=1;
for(long i=N;i>N-M;i--){
    ans=ans*i;
}
for(long i=M;i>0;i--){
    ans=ans/i;
}
return ans;
}

long Perm(long N,long M){// nPm
long ans=1;
for(long i=N;i>N-M;i--){
    ans=ans*i;
}
return ans;
}

long GCM(long A,long B){
            long W,r;
    if(A<B){

        W=B;
        B=A;
        A=W;
    }
    while(A%B!=0){
        r=A%B;
        A=B;
        B=r;
    }
    return B;
}
long LCM(long A,long B){
    return (A*B)/GCM(A,B);
}

long upDiv(long a,long b){
  return (a+b-1)/b;
}

long strtolong(string str){
istringstream ss;
ss=istringstream(str);
long num;
ss>>num;
return num;
}

long chartolong(char ch){
long num;
num=ch-'0';
return num;
}

string longtostr(long N){
    string str;
    str=to_string(N);
    return str;
}









/////////////////////////


//Answer


int main() {

long N,r,s;
cin>>N;
r=sqrt(N);
for(int i=r;i>0;i--){
    if(N%i==0){
        s=i;
        break;
    }
}
cout<<s+(N/s)-2<<endl;



}