Codeforces Round #180 (Div. 2) B. Sail 【模拟】

B. Sail

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The polar bears are going fishing. They plan to sail from (sx, sy) to (ex, ey). 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 for t seconds, what is the earliest time they sail to (ex, ey)?

Input

The first line contains five integers t, sx, sy, ex, ey (1 ≤ t ≤ 105,  - 109 ≤ sx, sy, ex, ey ≤ 109). The starting location and the ending location will be different.

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

Output

If they can reach (ex, ey) within t seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).

Examples

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.

原题链接：http://codeforces.com/problemset/problem/298/B

题意：一艘帆船要从一点到达另外一个点。每一秒钟都会有东南西北四个中的一种风，你可以选择动或不动，问你到达终点的最短时间。

直接模拟：如果某一时刻的风与你目标方向相同，则走，否则不动。

AC代码：

#include <bits/stdc++.h>using namespace std;const int maxn=1e5+5;char a[maxn];int main(){    int n,sx,sy,ex,ey;    while(cin>>n>>sx>>sy>>ex>>ey)    {        cin>>a;        bool flag=false;        for(int i=0;i<n;i++)        {            if(a[i]=='N'&&sy<ey)                sy++;            else if(a[i]=='S'&&sy>ey)                sy--;            else if(a[i]=='E'&&sx<ex)                sx++;            else if(a[i]=='W'&&sx>ex)                sx--;            if(sx==ex&&sy==ey)            {                flag=true;                cout<<i+1<<endl;                break;            }        }        if(!flag)        cout<<"-1"<<endl;    }    return 0;}