題幹：

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 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.

 

題目大意：

   給你定義四種操作，然後給你一個字串代表操作的順序，，當然，你可以選擇不去執行其中的某些操作。問你最終可否到達destination。（從一個座標到另一個座標）

解題報告：

   就是隨便模擬就行了，，，注意有個坑，，他對東南西北風的定義，，好像和中國人的 正好反著，，別忘讀題、、

AC程式碼：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e6 + 5;
int n,k;
char s[MAX],t[MAX];
int bk[550];
int cur[550];
int X1,X2,Y1,Y2;
//"E" (east), "S" (south), "W" (west) and "N" (north).
int main()
{
	cin>>t;
	cin>>X1>>Y1>>X2>>Y2;
	cin>>(s+1);
	int len = strlen(s+1);
	for(int i = 1; i<=len; i++) {
		bk[s[i]]++;
	} 
	int dx=X2-X1;
	int dy=Y2-Y1;
	int flag=0;
	int ans=0;
	if(dx>=0 && dy>=0) {
		if(bk['E'] >= dx && bk['N'] >= dy) {
			flag=1;
			for(int i = 1; i<=len; i++) {
				cur[s[i]]++;
				if(cur['E']>=dx && cur['N']>=dy) {
					ans=i;break;
				}
			}
		}
	}
	else if(dx<0 && dy>=0) {
		if(bk['W'] >= (-dx) && bk['N'] >= dy) {
			flag=1;
			for(int i = 1; i<=len; i++) {
				cur[s[i]]++;
				if(cur['W']>=(-dx) && cur['N']>=dy) {
					ans=i;break;
				}
			}			
		}
	}
	else if(dx<0 && dy<0) {
		if(bk['W'] >= (-dx) && bk['S'] >= (-dy)) {
			flag=1;
			for(int i = 1; i<=len; i++) {
				cur[s[i]]++;
				if(cur['W']>=(-dx) && cur['S']>=(-dy)) {
					ans=i;break;
				}
			}			
		}
	}
	else {
		if(bk['E'] >= dx && bk['S'] >= (-dy)) {
			flag=1;
			for(int i = 1; i<=len; i++) {
				cur[s[i]]++;
				if(cur['E']>=dx && cur['S']>=(-dy)) {
					ans=i;break;
				}
			}			
		}
	}
	if(flag) printf("%d\n",ans);
	else printf("-1\n");
	return 0 ;
 }
 /*
 5 0 0 1 1
SESNW
 
 */