5.29/B题

Description

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 Input

Input

2 1

Output

2

Input

10 7

Output

6

Input

1000000000000 1

Output

1000000000000

Hint

Pictures to the first and second sample test.

今天的B题，其实是很水的一道题，但就是没有1Y。

说下题目大意，输入两个数，a和b（a>=b），代表长和宽，然后开始折纸，问当折到不能再折后，总共折成了几块。

代码如下，十分简短

#include <cstdio>
long long cot=0;
long long ans(long long m,long long n)
{
    long long t;
    while (m!=n)
    {
        if (m<n) {t=m;m=n;n=t;}
        cot=cot+m/n;
        if (m%n==0) return cot;
        m=m%n;
    }
    return cot;
}
int main()
{
    long long a,b;
    scanf("%I64d%I64d",&a,&b);
    printf("%I64d\n",ans(a,b));
    return 0;
}