題干：

Amr loves Geometry. One day he came up with a very interesting problem.

Amr has a circle of radius?r?and center in point?(x,?y). He wants the circle center to be in new position?(x',?y').

In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.

Help Amr to achieve his goal in minimum number of steps.

Input

Input consists of 5 space-separated integers?r,?x,?y,?x'?y'?(1?≤?r?≤?105,??-?105?≤?x,?y,?x',?y'?≤?105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.

Output

Output a single integer — minimum number of steps required to move the center of the circle to the destination point.

Examples

Input

2 0 0 0 4

Output

1

Input

1 1 1 4 4

Output

3

Input

4 5 6 5 6

Output

0

Note

In the first sample test the optimal way is to put a pin at point?(0,?2)?and rotate the circle by?180?degrees counter-clockwise (or clockwise, no matter).

解題報告：

? ? ?不難發現，我們走一步可以走的最大距離就是走一個直徑，不難證明，小于一個直徑的距離，我們都可以通過調整旋轉點來一步到位，所以說啊，，直接貪心就出來了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5;int main() {ll r,x,xx,y,yy;cin>>r>>x>>y>>xx>>yy;r*=2;double dis = sqrt((xx-x)*(xx-x) + (yy-y)*(yy-y));if(x == xx && y == yy) puts("0");else {printf("%lld\n",(ll)ceil(dis/r));} return 0 ;}

?

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