A

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:

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).
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.
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.
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

Input

5
DDRRR

Output

Input

6
DDRRRR

Output

Note

Consider one of the voting scenarios for the first sample:

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

解题思路：仔细读一下题目，还是挺简单的。就是有两个门派D和R，给定一个字符串（只由D和R组成），只要位置靠前的人就可以将后面的对手deny掉，当轮到这个对手时，由于已被前面的人deny掉，所以此人再也没有deny他人的权利，直接跳过。不断循环，后面的人也可以deny前面还有选择权利的人，怎么实现呢？我们用队列来维护它们的位置（下标），当前面的人deny后面的对手后，此时就弹出两个队的队首元素（表示位置序号都已失效），但deny别人的人还有选择的权利，也可能会被后面的对手deny掉，于是只需将其位置序号加上n后入自己的队列中（满足位置序号比后面大，即后面的人可以deny掉"前面"还有选择权的对手）。最终，哪个队列不为空，谁就拥有vote的权利。

AC代码：

 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 }