Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 <= N <= 100,000) on a number line and the cow is at a point K (0 <= K <= 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting. * Walking: FJ can move from any point X to the points X-1 or X+1 in a single minute * Teleporting: FJ can move from any point X to the point 2*X in a single minute. If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

    農夫約翰被通知，他的一隻奶牛逃逸了！所以他決定，馬上幽發，儘快把那隻奶牛抓回來．     他們都站在數軸上．約翰在N(O≤N≤100000)處，奶牛在K(O≤K≤100000)處．約翰有 兩種辦法移動，步行和瞬移：步行每秒種可以讓約翰從z處走到x+l或x-l處；而瞬移則可讓他在1秒內從x處消失，在2x處出現．然而那隻逃逸的奶牛，悲劇地沒有發現自己的處境多麼糟糕，正站在那兒一動不動．     那麼，約翰需要多少時間抓住那隻牛呢？

Input

* Line 1: Two space-separated integers: N and K

    僅有兩個整數N和K.

 

Output

* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

    最短的時間．

Sample Input

5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.

Sample Output

4

OUTPUT DETAILS:

The fastest way for Farmer John to reach the fugitive cow is to
move along the following path: 5-10-9-18-17, which takes 4 minutes.     此題為寬搜