Codeforces-298b H Sail

B. Sail

time limit per test

1 second

memory limit per test

256 megabytes

The polar bears are going fishing. They plan to sail from (s_x, s_y) to(e_x, e_y). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (x, y).

• If the wind blows to the east, the boat will move to (x + 1, y).
• If the wind blows to the south, the boat will move to (x, y - 1).
• If the wind blows to the west, the boat will move to (x - 1, y).
• If the wind blows to the north, the boat will move to (x, y + 1).

Alternatively, they can hold the boat by the anchor. In this case, the boat stays at(x, y). Given the wind direction fort seconds, what is the earliest time they sail to(e_x, e_y)?

Input

The first line contains five integers t, s_x, s_y, e_x, e_y(1 ≤ t ≤ 10⁵,  - 10⁹ ≤ s_x, s_y, e_x, e_y ≤ 10⁹). The starting location and the ending location will be different.

The second line contains t characters, thei-th character is the wind blowing direction at thei-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).

Output

If they can reach (e_x, e_y) withint seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).

Sample test(s)

Input

5 0 0 1 1SESNW

Output

4

Input

10 5 3 3 6NENSWESNEE

Output

-1

Note

In the first sample, they can stay at seconds 1,3, and move at seconds 2,4.

In the second sample, they cannot sail to the destination.

——————————————————亚历山大的分割线——————————————————

思路：上高中的时候，数学老师教过一种小明从家里去学校有多少种路径的排列组合。。。咳咳，本题跟排列组合是木有关系滴！但是这个想法是要有的。横坐标差值对应了N或者S，纵坐标差值对应了W或者E。当前风向并不是期望的风向之一的话，等待一秒。

代码如下：

#include <stdio.h>#include <stdlib.h>#include <string.h>char dir[100010];char c, d;int sx, sy, ex, ey;void jud(){//jud()函数用来储存期望的风向，c表示东西，d表示南北    if(ex > sx&&ey > sy)        {d = 'N'; c = 'E';}else if(ex < sx&&ey > sy)        {d = 'N'; c = 'W';}else if(ex < sx&&ey < sy)        {d = 'S'; c = 'W';}    else if(ex > sx&&ey < sy)        {d = 'S'; c = 'E';}    else if(sx == ex)        {d = ey > sy ? 'N' : 'S'; c = ' ';}    else        {c = ex > sx ? 'E' : 'W'; d = ' ';}}int main(){int T, i, cou = 0;int stx, sty, step;scanf("%d%d%d%d%d", &T, &sx, &sy, &ex, &ey);scanf("%s", dir);stx = abs(sx - ex);//横坐标差值sty = abs(sy - ey);//纵坐标差值step = stx + sty;//一共需要的最少步数jud();for(i = 0; i < T; i++){        if(dir[i] != c&&dir[i] != d)  cou++;//cou的大小表示需要等待的时间        else if(dir[i] == c){            if(stx)  stx--;            else  cou++;         }        else if(dir[i] == d){            if(sty)  sty--;            else  cou++;        }        if(!stx && !sty)  break;}    if(stx || sty)        printf("-1\n");    else        printf("%d\n", cou + step);return 0;}