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).

Sample test(s)

input

5 0 0 1 1
SESNW

output

4

input

10 5 3 3 6
NENSWESNEE

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.

解题说明：此题就是按照题目要求进行模拟，每一次刮风就根据风向判断是否要改变位置，直到到达指定位置或者时间结束

#include<iostream>
#include<map>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;


int main()
{
	int t,a,b,x,y,i;
	int flag;
	char c;
	scanf("%d %d %d %d %d",&t,&a,&b,&x,&y);
	flag=0;
	for(i=1;i<=t;i++)
	{
		cin>>c;
		if(c=='N' && b<y)
		{
			b++;
		}
		else if(c=='S' && b>y)
		{
			b--;
		}
		else if(c=='E' && a<x)
		{
			a++;
		}
		else if(c=='W' && a>x)
		{
			a--;
		}
		if(a==x && b==y)
		{
			printf("%d\n",i);
			flag=1;
			break;
		}
	}
	if(flag==0)
	{
		printf("-1\n");
	}
}