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.

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.

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

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 ;

}