527A. Playing with Paper【math】

A. Playing with Paper

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular a mm  ×  b mm sheet of paper (a > b). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.

After making a paper ship from the square piece, Vasya looked on the remaining (a - b) mm  ×  b mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.

Can you determine how many ships Vasya will make during the lesson?

Input

The first line of the input contains two integers a, b (1 ≤ b < a ≤ 10¹²) — the sizes of the original sheet of paper.

Output

Print a single integer — the number of ships that Vasya will make.

Sample test(s)

Input

2 1

Output

2

Input

10 7

Output

6

Input

1000000000000 1

Output

1000000000000

Note

Pictures to the first and second sample test.

题意：一张纸，居短边对折直到最后一块是正方形。问把原先的矩形分成了多少块。 分析：如果x，y是两边，x比y大2倍以上的话就会一直顺着y边对折，几个二倍关系就折几次，找出这层关系即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<algorithm>
#include <map>
#include <queue>
#include <set>
using namespace std;

//2015.3.22  527A  10的12次幂
long long int n,m;
int main()
{
    cin>>n>>m;
    long long int step=0;
    while(n&&m)
    {
        step+=n/m;
        n%=m;
        swap(n,m);
    }
    cout<<step;
    return 0;
}