Codeforces Round #340 (Div. 2)-C. Watering Flowers

原题链接

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 xi and yi ( - 107 ≤ xi, yi ≤ 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.

Examples

input

2 -1 0 5 30 25 2

output

6

input

4 0 0 5 09 48 3-1 01 4

output

33

算出每一朵花到第一个喷泉的距离，再把每朵花按照距离排序，从大到小遍历每个距离，当作第一个喷泉的r1^2, 在算出剩下的花和第二个喷泉的距离，取最大值，当作第二个喷泉的r2^2

#include <bits/stdc++.h>#define maxn 2005using namespace std;typedef long long ll;struct Node{bool friend operator < (const Node&a, const Node&b){return a.d1 < b.d1;}ll x, y;ll d1, d2;}node[maxn];int main(){//freopen("in.txt", "r", stdin);int n;ll x1, y1, x2, y2;scanf("%d%I64d%I64d%I64d%I64d", &n, &x1, &y1, &x2, &y2);for(int i = 1; i <= n; i++){ scanf("%I64d%I64d", &node[i].x, &node[i].y); node[i].d1 = (node[i].x - x1) * (node[i].x - x1) + (node[i].y - y1) * (node[i].y - y1);    }    sort(node+1, node+n+1);    ll ans = 1e18;    for(int i = n; i >= 0; i--){    node[i].d2 = (node[i].x - x2) * (node[i].x - x2) + (node[i].y - y2) * (node[i].y - y2);    node[i].d2 = max(node[i].d2, node[i+1].d2);    ans = min(ans, node[i].d1 + node[i+1].d2);    }    printf("%I64d\n", ans);        return 0;}