Codeforces Round #370 (Div. 2)C. Memory and De-Evolution 贪心

地址：http://codeforces.com/problemset/problem/712/C

题目：

C. Memory and De-Evolution

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

input

6 3

output

4

input

8 5

output

3

input

22 4

output

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: 

.

 

 思路：倒着贪心就好，每次把最小的边变换成最大的。。

 

代码：

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int, int> PII;
const double eps = 1e-8;
const double pi = acos(-1.0);
const int K = 1e6 + 7;
int a[4];
int main(void)
{
    int ans=0,st,se;
    cin>>st>>se;
    a[1]=a[2]=a[3]=se;
    while(1)
    {
        sort(a+1,a+1+3);
        if(a[1]<st)
            a[1]=min(a[2]+a[3]-1,st);
        else
            break;
        //printf("%d:%d %d %d\n",ans+1,a[1],a[2],a[3]);
        ans++;
    }
    cout<<ans<<endl;
    return 0;
}