#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
#define INF 0x3f3f3f
char a[100050];
char b[100050];
int n;
int solve(int p,int q,int sum1,int sum2)
{while(q!=n){if(b[q]==b[p]&&b[p]=='r'){b[q]='b';sum1++;p++;q++;}else if(b[q]==b[p]&&b[p]=='b'){b[q]='r';sum2++;p++;q++;}else{p++;q++;}}return max(sum1,sum2);//注意这里是什么意思，sum1代表r需要改变的次数，sum2代表b需要改变的次数，所以可以                              //交换的次数就是min(sum1,sum2),而剩下的就必须是改变的，大的减去小的加上min(sum1,sum2)就等价于max(sum1,sum2)
}
int main()
{int ans1,ans2;//sum1是b的转移次数，sum2是r的转移次数cin>>n;cin>>a;strcpy(b,a);//cout<<b;if(b[0]=='r'){ans1=solve(0,1,0,0);strcpy(b,a);b[0]='b';ans2=solve(0,1,1,0);}else if(b[0]=='b'){ans1=solve(0,1,0,0);strcpy(b,a);b[0]='r';ans2=solve(0,1,0,1);}// cout<<sum1<<sum2<<sum3<<sum4<<endl;cout<<min(ans1,ans2);return 0;
}

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?

Input

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.

Output

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.

Examples

6 3

4

8 5

3

22 4

6

Note

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:

.

题目大意：输入x,y，x代表刚开始的等边三角形的边长，y代表最终的等边三角形的边长，问你从x变为y,最少要变多少次，注意，变的过程中也要是三角形

个人思路：刚开始在找规律，应该说找到一个错误的规律，所以wa了，然后没什么想法了，百度了一下，看到个逆序来求，然后瞬间把它秒了，就是逆序来做

看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
const int maxn=1e5+10;
const ll maxa=1e10;
#define INF 0x3f3f3f3f3f3f
int a[5];
int x,y,ans=0;
void solve()
{sort(a,a+3);int t=a[1]+a[2]-1;if(t<=x){a[0]=t;ans++;}else{a[0]=x;ans++;}
}
int main()
{cin>>x>>y;for(int i=0;i<3;i++)a[i]=y;while(a[0]!=x||a[1]!=x||a[2]!=x){solve();}cout<<ans<<endl;return 0;
}