（CodeForces

（CodeForces - 598C）Nearest vectors

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.

Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 and π. For example, opposite directions vectors have angle equals to π.

Input

First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.

The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed that no two vectors in the input share the same direction (but they still can have opposite directions).

Output

Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any.

Examples

Input

4
-1 0
0 -1
1 0
1 1

Output

3 4

Input

6
-1 0
0 -1
1 0
1 1
-4 -5
-4 -6

Output

6 5

题目大意：给出二维平面的一系列向量，这些向量的起点都在原点，问哪两个向量的夹角最小。

思路：根据向量与x轴的夹角排序，之后一一枚举，找出夹角最小的两个向量即可。
**ps：这题的精度很高可以使用long double，精度eps控制在1e-20左右；C语言里long double atan2(long double y,long double x) 返回的是原点至点(x,y)的方位角，即与 x 轴的夹角，也可以理解为复数 x+yi 的辐角，返回值的单位为弧度，取值范围为(-PI,PI].**long double提交的时候要选高版本的G++。

#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const long double pi=acos(-1.0);const int maxn=100005;struct node{    long double x,y;    long double t;    int id;}a[maxn];bool cmp(node a,node b){    return a.t<b.t;}int main(){    int n;    scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%Lf%Lf",&a[i].x,&a[i].y);        a[i].id=i+1;        a[i].t=atan2(a[i].y,a[i].x);    }    sort(a,a+n,cmp);    long double m=2*pi;    int ans;    for(int i=0;i<n-1;i++)        if((a[i+1].t-a[i].t)<m)        {            ans=i;            m=a[i+1].t-a[i].t;                  }    if((2*pi-(a[n-1].t-a[0].t))<m) printf("%d %d\n",a[0].id,a[n-1].id);    else printf("%d %d\n",a[ans].id,a[ans+1].id);    return 0;}