Rng is baking cookies.

Initially, he can bake one cookie per second.

He can also eat the cookies baked by himself. When there are x cookies not yet eaten, he can choose to eat all those cookies. After he finishes eating those cookies, the number of cookies he can bake per second becomes x. Note that a cookie always needs to be baked for 1 second, that is, he cannot bake a cookie in 1/x seconds when x > 1. When he choose to eat the cookies, he must eat all of them; he cannot choose to eat only part of them. It takes him A seconds to eat the cookies regardless of how many, during which no cookies can be baked.

He wants to give N cookies to Grandma. Find the shortest time needed to produce at least N cookies not yet eaten.

Constraints

* 1≦N≦10^{12}
* 0≦A≦10^{12}
* A is an integer.

Input

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


N A


Output

Print the shortest time needed to produce at least N cookies not yet eaten.

Examples

Input

8 1


Output

7


Input

1000000000000 1000000000000


Output

1000000000000
#include <bits/stdc++.h>
using namespace std;

#define FOR(i,a,b) for(int i=a;i<b;i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back
#define REACH cerr<<"reached line "<<__LINE__<<endl
#define DBG(x) cerr<<"line "<<__LINE__<<" "<<#x<<":"<<x<<endl

using uint=unsigned int;
using ll=long long;
using pii=pair<int,int>;
using vi=vector<int>;
using ld=long double;

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"[";
	REP(i,(int)v.size()){
		if(i)os<<",";
		os<<v[i];
	}
	os<<"]";
	return os;
}

int read(){
	int i;
	scanf("%d",&i);
	return i;
}

ll readLL(){
	ll i;
	scanf("%lld",&i);
	return i;
}

string readString(){
	static char buf[3341919];
	scanf("%s",buf);
	return string(buf);
}

char* readCharArray(){
	static char buf[3341919];
	static int bufUsed=0;
	char* ret=buf+bufUsed;
	scanf("%s",ret);
	bufUsed+=strlen(ret)+1;
	return ret;
}

template<class T,class U>
void chmax(T& a,U b){
	if(a<b)
		a=b;
}

template<class T,class U>
void chmin(T& a,U b){
	if(a>b)
		a=b;
}

template<class T>
T Sq(const T& t){
	return t*t;
}

const int mod=1000000007;
template<class T,class U>
void add(T& a,U b){
	a=((ll)a+b)%mod;
}

template<>
void add(int& a,int b){
	a+=b;
	if(a>=mod)a-=mod;
}

template<class T,class U>
void sub(T& a,U b){
	a=((ll)a-b%mod+mod)%mod;
}

template<>
void sub(int& a,int b){
	a-=b;
	if(a<0)a+=mod;
}

template<class T,class U>
void mult(T& a,U b){
	a=((ll)a*b)%mod;
}

const int S=3341919;
using int128=__int128;
struct Line{
	int128 a,b;
	int128 calc(int128 x){
		return a*x+b;
	}
};
struct ConvexHull{
	Line buf[S];
	int b,e;
	void Init(){
		b=e=0;
	}
	void Push(Line l){
		while(e-b>=2){
			Line p=buf[e-2],q=buf[e-1];
			if(l.a*(p.b-q.b)+l.b*(q.a-p.a)>=q.a*p.b-p.a*q.b)
				e--;
			else
				break;
		}
		buf[e++]=l;
	}
	int128 Get(int128 x){
		while(e-b>=2){
			if(buf[b].calc(x)<=buf[b+1].calc(x))
				b++;
			else
				break;
		}
		return buf[b].calc(x);
	}
} ch;

int main(){
	ll n=readLL(),a=readLL();
	ch.Init();
	ch.Push(Line{int128(1),int128(0)});
	ll ans=n;
	FOR(i,2,S){
		int128 v=ch.Get(i);
		if(v>=n){
			chmin(ans,i);
			break;
		}
		ch.Push(Line{v,-v*(a+i)});
		chmin(ans,a+i+(n-1)/v+1);
	}
	cout<<ans<<endl;
}
