codeforces_672C. Recycling Bottles（贪心）

C. Recycling Bottles

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

It was recycling day in Kekoland. To celebrate it Adil and Bera went to Central Perk where they can take bottles from the ground and put them into a recycling bin.

We can think Central Perk as coordinate plane. There are n bottles on the ground, the i-th bottle is located at position (xi, yi). Both Adil and Bera can carry only one bottle at once each.

For both Adil and Bera the process looks as follows:

1. Choose to stop or to continue to collect bottles.
2. If the choice was to continue then choose some bottle and walk towards it.
3. Pick this bottle and walk to the recycling bin.
4. Go to step 1.

Adil and Bera may move independently. They are allowed to pick bottles simultaneously, all bottles may be picked by any of the two, it's allowed that one of them stays still while the other one continues to pick bottles.

They want to organize the process such that the total distance they walk (the sum of distance walked by Adil and distance walked by Bera) is minimum possible. Of course, at the end all bottles should lie in the recycling bin.

Input

First line of the input contains six integers ax, ay, bx, by, tx and ty (0 ≤ ax, ay, bx, by, tx, ty ≤ 109) — initial positions of Adil, Bera and recycling bin respectively.

The second line contains a single integer n (1 ≤ n ≤ 100 000) — the number of bottles on the ground.

Then follow n lines, each of them contains two integers xi and yi (0 ≤ xi, yi ≤ 109) — position of the i-th bottle.

It's guaranteed that positions of Adil, Bera, recycling bin and all bottles are distinct.

Output

Print one real number — the minimum possible total distance Adil and Bera need to walk in order to put all bottles into recycling bin. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if .

Examples

input

3 1 1 2 0 031 12 12 3

output

11.084259940083

input

5 0 4 2 2 055 23 05 53 53 3

output

33.121375178000

Note

Consider the first sample.

Adil will use the following path: .

Bera will use the following path: .

Adil's path will be  units long, while Bera's path will be  units long.

分四种情况

1、A先取，B后取。

2、B先取，A后取

3、只有A取 

4、只有B取

找出Bera或Adil到垃圾桶的距离与Bera或Adil到瓶子的距离的差值最大的那个瓶子，

这样在总距离上就减少了这个差值的大小。注意差值也可能全是负的。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define Si(a) scanf("%d",&a)#define Sl(a) scanf("%lld",&a)#define Sd(a) scanf("%lf",&a)#define Ss(a) scanf("%s",a)#define Pi(a) printf("%d\n",(a))#define Pl(a) printf("%lld\n",(a))#define Pd(a) printf("%lf\n",(a))#define Ps(a) printf("%s\n",(a))#define W(a) while(a--)#define mem(a,b) memset(a,(b),sizeof(a))#define FOP freopen("data.txt","r",stdin)#define inf 1e30#define maxn 1000010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;struct p{    double x,y,la,lb,lt;}bot[100010];double solve(double ax,double ay,double bx,double by){    return sqrt((ax-bx)*(ax-bx)+(ay-by)*(ay-by));}int main(){    double ax,ay,bx,by,tx,ty;    int i,n;    scanf("%lf%lf%lf%lf%lf%lf",&ax,&ay,&bx,&by,&tx,&ty);    Si(n);    for(i=1;i<=n;i++)    {        Sd(bot[i].x),Sd(bot[i].y);        bot[i].lt=solve(bot[i].x,bot[i].y,tx,ty);        bot[i].la=solve(bot[i].x,bot[i].y,ax,ay);        bot[i].lb=solve(bot[i].x,bot[i].y,bx,by);    }    int mai=0,mbi=0;    double ma=-inf,mb=-inf;    double sum1=0,sum2=0,sum3=0,sum4=0;    for(i=1;i<=n;i++)    {        if(bot[i].lt-bot[i].la>ma)        {            ma=bot[i].lt-bot[i].la;            mai=i;        }    }    for(i=1;i<=n;i++)    {        if(i==mai)continue;        if(bot[i].lt-bot[i].lb>mb)        {            mb=bot[i].lt-bot[i].lb;            mbi=i;        }    }    sum1=bot[mai].la+bot[mai].lt+bot[mbi].lb+bot[mbi].lt;    for(i=1;i<=n;i++)    {        if(i==mai||i==mbi)continue;        sum1+=bot[i].lt*2;    }    ma=-inf;mb=-inf;mai=0;mbi=0;    for(i=1;i<=n;i++)    {        if(bot[i].lt-bot[i].lb>mb)        {            mb=bot[i].lt-bot[i].lb;            mbi=i;        }    }    for(i=1;i<=n;i++)    {        if(i==mbi)continue;        if(bot[i].lt-bot[i].la>ma)        {            ma=bot[i].lt-bot[i].la;            mai=i;        }    }    sum2=bot[mai].la+bot[mai].lt+bot[mbi].lb+bot[mbi].lt;    for(i=1;i<=n;i++)    {        if(i==mai||i==mbi)continue;        sum2+=bot[i].lt*2;    }    ma=-inf;mb=-inf;mai=0;mbi=0;    for(i=1;i<=n;i++)    {        if(bot[i].lt-bot[i].la>ma)        {            ma=bot[i].lt-bot[i].la;            mai=i;        }    }    sum3=bot[mai].la+bot[mai].lt;    for(i=1;i<=n;i++)    {        if(i==mai)continue;        sum3+=bot[i].lt*2;    }    ma=-inf;mb=-inf;mai=0;mbi=0;    for(i=1;i<=n;i++)    {        if(bot[i].lt-bot[i].lb>mb)        {            mb=bot[i].lt-bot[i].lb;            mbi=i;        }    }    sum4=bot[mbi].lb+bot[mbi].lt;    for(i=1;i<=n;i++)    {        if(i==mbi)continue;        sum4+=bot[i].lt*2;    }    printf("%.12lf\n",min(min(sum1,sum2),min(sum3,sum4)));    return 0;}