cf340 C. Watering Flowers

C. Watering Flowers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A flowerbed has many flowers and two fountains.

You can adjust the water pressure and set any values r₁(r₁ ≥ 0) and r₂(r₂ ≥ 0), giving the distances at which the water is spread from the first and second fountain respectively. You have to set such r₁ and r₂ that all the flowers are watered, that is, for each flower, the distance between the flower and the first fountain doesn't exceed r₁, or the distance to the second fountain doesn't exceed r₂. It's OK if some flowers are watered by both fountains.

You need to decrease the amount of water you need, that is set such r₁ and r₂ that all the flowers are watered and the r₁² + r₂² is minimum possible. Find this minimum value.

Input

The first line of the input contains integers n, x₁, y₁, x₂, y₂ (1 ≤ n ≤ 2000,  - 10⁷ ≤ x₁, y₁, x₂, y₂ ≤ 10⁷) — the number of flowers, the coordinates of the first and the second fountain.

Next follow n lines. The i-th of these lines contains integers x_i and y_i ( - 10⁷ ≤ x_i, y_i ≤ 10⁷) — the coordinates of the i-th flower.

It is guaranteed that all n + 2 points in the input are distinct.

Output

Print the minimum possible value r₁² + r₂². Note, that in this problem optimal answer is always integer.

Sample test(s)

Input

2 -1 0 5 3
0 2
5 2

Output

6

Input

4 0 0 5 0
9 4
8 3
-1 0
1 4

Output

33

Note

The first sample is (r₁² = 5, r₂² = 1): The second sample is (r₁² = 1, r₂² = 32):

思路：直接枚举就可以了，r1可以是0或者是到其他任意一个点的距离，然后就枚举r2,条件是到1的距离大于r1中最大，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const long long INF = 10e18;
const int MAX =  + ;
long long Distance1[MAX],Distance2[MAX];
long long get_distance(long long x1,long long y1,long long x2, long long y2)
{
    return (x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2);
}
int main()
{
    int n;
    long long x1,x2,y1,y2,x,y;
    long long r1 = ,r2 = ;
    long long ans = INF;
    scanf("%d%I64d%I64d%I64d%I64d", &n,&x1,&y1,&x2,&y2);
    Distance1[] = Distance2[] = ;
    for(int i = ; i <= n; i++)
    {
        scanf("%I64d%I64d",&x,&y);
        Distance1[i] = get_distance(x1,y1,x,y);
        Distance2[i] = get_distance(x2,y2,x,y);
    }

    for(int i = ; i <= n; i++)
    {
        r1 = Distance1[i];
        r2 = ;
        for(int j = ; j <= n; j++)
        {
            if(Distance1[j] > r1 && Distance2[j] > r2)
            {
                r2 = Distance2[j];
            }
        }
        ans = min(ans, r1 + r2);
    }

    printf("%I64d\n",ans);
    return ;
}