【Wunder Fund Round 2016 (Div 1 + Div 2 combined)C】【排序 共线判定】平面n点不共一线，寻找不含点的任一三角形

C. Constellation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Cat Noku has obtained a map of the night sky. On this map, he found a constellation with n stars numbered from 1 to n. For each i, thei-th star is located at coordinates (x_i, y_i). No two stars are located at the same position.

In the evening Noku is going to take a look at the night sky. He would like to find three distinct stars and form a triangle. The triangle must have positive area. In addition, all other stars must lie strictly outside of this triangle. He is having trouble finding the answer and would like your help. Your job is to find the indices of three stars that would form a triangle that satisfies all the conditions.

It is guaranteed that there is no line such that all stars lie on that line. It can be proven that if the previous condition is satisfied, there exists a solution to this problem.

Input

The first line of the input contains a single integer n (3 ≤ n ≤ 100 000).

Each of the next n lines contains two integers x_i and y_i ( - 10⁹ ≤ x_i, y_i ≤ 10⁹).

It is guaranteed that no two stars lie at the same point, and there does not exist a line such that all stars lie on that line.

Output

Print three distinct integers on a single line — the indices of the three points that form a triangle that satisfies the conditions stated in the problem.

If there are multiple possible answers, you may print any of them.

Examples

input

3
0 1
1 0
1 1

output

1 2 3

input

5
0 0
0 2
2 0
2 2
1 1

output

1 3 5

Note

In the first sample, we can print the three indices in any order.

In the second sample, we have the following picture.

Note that the triangle formed by starts 1, 4 and 3 doesn't satisfy the conditions stated in the problem, as point 5 is not strictly outside of this triangle (it lies on it's border).

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 1e5+10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
struct A
{
	int x, y, o;
	bool operator < (const A& b)const
	{
		if (x != b.x)return x < b.x;
		if (y != b.y)return y < b.y;
	}
}a[N];
int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d%d", &a[i].x, &a[i].y);
			a[i].o = i;
		}
		sort(a + 1, a + n + 1);
		for (int i = 3; i <= n; ++i)
		{
			LL x1 = a[i].x - a[1].x;
			LL y1 = a[i].y - a[1].y;
			LL x2 = a[i].x - a[2].x;
			LL y2 = a[i].y - a[2].y;
			if (x1*y2 != x2*y1)
			{
				printf("%d %d %d\n", a[1].o, a[2].o, a[i].o);
				break;
			}
		}
	}
	return 0;
}
/*
【题意】
平面上给你n（3<=n<=1e5）个点，这n个点不共线。
让你找出3个点，要满足这3个点内不能有其他任何点。

【类型】
暴力

【分析】
我们按照先x后y排序，找出前2个点，
然后从第3点开始枚举，枚举到的第一个不共线的点，就是第3点。
答案就出来啦。

【时间复杂度&&优化】
O(nlogn)

*/