Ada the Ladybug has many things to do. She puts them into her queue. Anyway she is very indecisive, so sometime she uses the top, sometime the back and sometime she decides to reverses it.

Input

The first line consists of?1 ≤ Q ≤ 10⁶, number of queries. Each of them contains one of following commands

back - Print number from back and then erase it

front - Print number from front and then erase it

reverse - Reverses all elements in queue

push_back?N?- Add element?N?to back

toFront?N?- Put element?N?to front

All numbers will be?0 ≤ N ≤ 100

Output

For each back/front query print appropriate number.

If you would get this type of query and the queue would be empty, print "No job for Ada?" instead.

Example Input

15 toFront 93 front back reverse back reverse toFront 80 push_back 53 push_back 50 front front reverse push_back 66 reverse front

Example Output

93 No job for Ada? No job for Ada? 80 53

66

解析：雙向隊列模擬；主要是reverse,定義一個flag標(biāo)記，對頭隊尾即可；

#include<bits/stdc++.h> using namespace std; int Q;char s[15]; int num; deque<int>q; deque<int>qq;deque<int>::iterator it;int main() {int ans=0;scanf("%d",&Q);for(int i=0;i<Q;i++){scanf("%s",s);if(s[0]=='t'){scanf("%d",&num);if(ans%2==0)q.push_front(num);elseq.push_back(num);}else if(s[0]=='p'){scanf("%d",&num);if(ans%2==0)q.push_back(num);elseq.push_front(num);}else if(s[0]=='f'){if(q.empty()){puts("No job for Ada?");}else{if(ans%2==0){int top=q.front();q.pop_front();printf("%d\n",top);}else{int tail=q.back();q.pop_back();printf("%d\n",tail);}}}else if(s[0]=='b'){if(q.empty()){puts("No job for Ada?");}else{if(ans%2==0){int tail=q.back();q.pop_back();printf("%d\n",tail);}else{int top=q.front();q.pop_front();printf("%d\n",top);}}}else{ans++;}} }

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