有幾個需要注意的地方：1，不用考慮在一秒內，兩個兔子移動會不會相遇，如果考慮就真的難了； 2，相遇的時候，如果到了左轉的時間，會優先處理相遇的情況；3，碰到牆反轉方向的時候，不計時；

小技巧：用編號表示方向，編號+2取餘就是反轉的方向編號，+3取餘就是左轉的方向編號，+1取餘是右轉的；

一秒前進的位置最好是一格一格的模擬，碰到牆就反轉方向，不用擔心超時

注意：我寫的程式碼的xy軸和題目是相反的，題目的x軸是縱向的，我的是數學的那種標準；

特別麻煩的題目；

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<sstream>
#include<vector>
#include<string>
#include<set>

using namespace std;
//using namespace __gnu_pbds;

#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define REP(i,n) for(int i=0;i<n;++i)

int read(){

    int r=0,f=1;char p=getchar();
    while(p>'9'||p<'0'){if(p=='-')f=-1;p=getchar();}
    while(p>='0'&&p<='9'){r=r*10+p-48;p=getchar();}return r*f;
}

//typedef tree<pair<long long,int>,null_type,less< pair<long long,int> >,rb_tree_tag,tree_order_statistics_node_update> rbtree;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
const int Maxn = 1e5+10;
const long long LINF = 1e18;
const int INF = 0x3f3f3f3f;
const int Mod = 10001;
const double PI = acos(-1.0);

struct Node {
    int x,y;
} dir[4];

map<string, int> mp;
map<int,string> mp2;

bool check (int &x, int &y,int n) {
    if(x < 1 || x > n || y < 1 || y > n) {
        x = x < 1 ? 1 : x;
        x = x > n ? n : x;
        y = y < 1 ? 1 : y;
        y = y > n ? n : y;
        return true;
    }
    else return false;
}

void chage (string &a, string &b) { //　相遇
    string tmp = a;
    a = b; b = tmp;
}

void chage2 (string &a,int sg) { // 1掉頭，2左轉
    if(sg == 1) a = mp2[(mp[a]+2)%4];
    else a = mp2[(mp[a]+3)%4];
}

int main (void)
{
    IOS;
    int N;
    mp["N"] = 0; mp["E"] = 1; mp["S"] = 2; mp["W"] = 3;
    mp2[0] = "N"; mp2[1] = "E"; mp2[2] = "S"; mp2[3] = "W";
    dir[0].x = 0; dir[0].y = -1; dir[1].x = 1; dir[1].y = 0;
    dir[2].x = 0; dir[2].y = 1; dir[3].x = -1; dir[3].y = 0;
	while (cin >> N) {
        if(!N) break;
        string d1,d2; int sp1,sp2,t1,t2,x1,y1,x2,y2;
        x1 = y1 = 1; x2 = y2 = N;
        cin >> d1 >> sp1 >> t1;
        cin >> d2 >> sp2 >> t2;
        int p,cnt = 0; cin >> p;
        while (++cnt <= p) {
            int tmp = 0;
            while (++tmp <= sp1) {
                x1+=dir[mp[d1]].x; y1+=dir[mp[d1]].y;
                if(check(x1,y1,N)) {
                    chage2(d1,1); --tmp;
                }
            }
            tmp = 0;
            while (++tmp <= sp2) {
                x2+=dir[mp[d2]].x; y2+=dir[mp[d2]].y;
                if(check(x2,y2,N)) {
                    chage2(d2,1); --tmp;
                }
            }
            if(x1 == x2 && y1 == y2) chage (d1,d2);
            else {
                if(cnt%t1 == 0) chage2(d1,2);
                if(cnt%t2 == 0) chage2(d2,2);
            }
        }
        cout << y1 << " " << x1 << endl;
        cout << y2 << " " << x2 << endl;
 	}
   	return 0;
}