Problem description

There are n employees in Alternative Cake Manufacturing (ACM). They are now voting on some very important question and the leading world media are trying to predict the outcome of the vote.

Each of the employees belongs to one of two fractions: depublicans or remocrats, and these two fractions have opposite opinions on what should be the outcome of the vote. The voting procedure is rather complicated:

1. Each of n employees makes a statement. They make statements one by one starting from employees 1 and finishing with employee n. If at the moment when it‘s time for the i-th employee to make a statement he no longer has the right to vote, he just skips his turn (and no longer takes part in this voting).
2. When employee makes a statement, he can do nothing or declare that one of the other employees no longer has a right to vote. It‘s allowed to deny from voting people who already made the statement or people who are only waiting to do so. If someone is denied from voting he no longer participates in the voting till the very end.
3. When all employees are done with their statements, the procedure repeats: again, each employees starting from 1 and finishing with n who are still eligible to vote make their statements.
4. The process repeats until there is only one employee eligible to vote remaining and he determines the outcome of the whole voting. Of course, he votes for the decision suitable for his fraction.

You know the order employees are going to vote and that they behave optimal (and they also know the order and who belongs to which fraction). Predict the outcome of the vote.

Input

The first line of the input contains a single integer n (1?≤?n?≤?200?000) — the number of employees.

The next line contains n characters. The i-th character is ‘D‘ if the i-th employee is from depublicans fraction or ‘R‘ if he is from remocrats.

Output

Print ‘D‘ if the outcome of the vote will be suitable for depublicans and ‘R‘ if remocrats will win.

Examples

5
DDRRR

D

6
DDRRRR

R

Note

Consider one of the voting scenarios for the first sample:

1. Employee 1 denies employee 5 to vote.
2. Employee 2 denies employee 3 to vote.
3. Employee 3 has no right to vote and skips his turn (he was denied by employee 2).
4. Employee 4 denies employee 2 to vote.
5. Employee 5 has no right to vote and skips his turn (he was denied by employee 1).
6. Employee 1 denies employee 4.
7. Only employee 1 now has the right to vote so the voting ends with the victory of depublicans.

解題思路：仔細讀一下題目，還是挺簡單的。就是有兩個門派D和R，給定一個字符串（只由D和R組成），只要位置靠前的人就可以將後面的對手deny掉，當輪到這個對手時，由於已被前面的人deny掉，所以此人再也沒有deny他人的權利，直接跳過。不斷循環，後面的人也可以deny前面還有選擇權利的人，怎麽實現呢？我們用隊列來維護它們的位置（下標），當前面的人deny後面的對手後，此時同時出兩個隊的隊首元素（表示位置上的人已經失效），但deny別人的人可能會被後面的對手deny掉，於是只需將其位置序號加上n後入自己的隊列中（滿足位置序號比後面大，即後面的人可以deny掉前面還有選擇權的對手）。最終，哪個隊列不為空，誰就擁有vote的權利。

AC代碼：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n;char s[200005];
 4 queue<int> d,r;
 5 int main(){
 6     cin>>n;getchar();
 7     cin>>s;
 8     for(int i=0;i<n;++i){
 9         if(s[i]==‘D‘)d.push(i);
10         else r.push(i);
11     }
12     while(!d.empty()&&!r.empty()){
13         int dd=d.front(),rr=r.front();
14         if(dd<rr){d.pop();r.pop();d.push(n+dd);}
15         else{d.pop();r.pop();r.push(n+rr);}
16     }
17     if(!d.empty())cout<<‘D‘<<endl;
18     else cout<<‘R‘<<endl;
19     return 0;
20 }

A - Voting