CodeForces

Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game hasn clock stations, station number i is at point (x_i, y_i) of the plane. As the player visits station numberi, he increases the current time on his timer bya_i. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.

A player spends d·dist time units to move between stations, wheredist is the distance the player has covered andd is some constant. The distance between stationsi and j is determined as|x_i - x_j| + |y_i - y_j|.

Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).

Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.

Input

The first line contains integers n and d (3 ≤ n ≤ 100, 10³ ≤ d ≤ 10⁵) — the number of stations and the constant from the statement.

The second line contains n - 2 integers:a₂, a₃, ..., a_n - 1(1 ≤ a_i ≤ 10³). The nextn lines contain the coordinates of the stations. Thei-th of them contains two integers x_i, y_i (-100 ≤ x_i, y_i ≤ 100).

It is guaranteed that no two stations are located at the same point.

Output

In a single line print an integer — the answer to the problem.

Example

Input

3 100010000 00 10 3

Output

2000

Input

3 100010001 01 11 2

Output

1000

题意：给你n个点，给你d的值，然后给你2~n-1的补给值，再后面是每个点的坐标，问你从1到n的最时间是多少。两点的时间是d* |x_i - x_j| + |y_i - y_j|..

思路：将n个点的转变成地图，Map[i][j]存的是i到j的时间 。

#include<iostream>#include<cstdio>#include<functional>#include<queue>#include<cmath>#include<cstring>using namespace std;#define N 99999999int n,d;int Map[105][105];int v[105];struct node {    int x;    int y;} a[105];int dis(node a,node b) {    return abs(a.x-b.x)+abs(a.y-b.y);}int main() {    int i,j,k;    while(scanf("%d%d",&n,&d)!=EOF) {        memset(v,0,sizeof(v));        for(i=2; i<n; i++)            scanf("%d",&v[i]);//每个点的补给值               for(i=0;i<=n;i++){            for(j=0;j<=n;j++){                if(i==j) Map[i][j]=0;                else Map[i][j]=N;//对地图初始化            }        }         for(i=1; i<=n; i++) {            scanf("%d%d",&a[i].x,&a[i].y);//存一下每个点的坐标        }        for(i=1; i<=n; i++) {            for(j=1; j<=n; j++) {                if(i==j) continue;                Map[i][j]=d*dis(a[i],a[j])-v[j];//将坐标转化地图             //   printf("d=%d dis=%lf\n",d,dis(a[i].x,a[i].y,a[j].x,a[j].y));            //    printf("i=%d j=%d Map=%lf\n",i,j,Map[i][j]);            }        }        /*        for(i=1;i<=n;i++){            for(j=1;j<=n;j++){                printf("%d ",Map[i][j]);            }            printf("\n");        }        */            for(k=1; k<=n; k++) {                for(i=1; i<=n; i++) {                    for(j=1; j<=n; j++) {                        // printf("i=%d j=%d k=%d\n",i,j,k);                        // printf("M[i][j]=%d  M[i][k]=%d  M[k][j]=%d v[k]=%d\n",Map[i][j],Map[i][k],Map[k][j],v[k]);                        if(i==j)                        continue;                        if(Map[i][j]>(Map[i][k]+Map[k][j]))                            Map[i][j]=Map[i][k]+Map[k][j];//弗洛伊德                        //   printf("Map[i][j]=%d\n",Map[i][j]);                    }                }            }            printf("%ld\n",Map[1][n]);//输出1到n的值    }}