Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
The first and only line contains two integers x and y (3 ≤ y < x ≤ 100 000) — the starting and ending equilateral triangle side lengths respectively.
Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
6 3
4
8 5
3
22 4
6
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do 
.
In the second sample test, Memory can do 
.
In the third sample test, Memory can do: 
.
简单贪心，逆着推，满足三角形的两边和大雨第三边
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
ll a[3];
ll n,m;
ll re ;
int main(int argc, char** argv)
{
cin >> n >> m;
a[0] = m;
a[1] = m;
a[2] = m;
while(!(a[0]==n&&a[1]==n&&a[2]==n))
{
sort(a,a+3);
// cout << a[0] << " "<< a[1] << " "<< a[2] <<" "<<endl;
ll t = a[1]+a[2];
if(t-1>n)
{
a[0] = n;
}
else
a[0] = t-1;
re++;
}
cout << re <<endl;
return 0;
}