Problem:
Score : 1400 points

Problem Statement

Takahashi found an integer sequence (A_1,A_2,...,A_N) with N terms. Since it was too heavy to carry, he decided to compress it into a single integer.

The compression takes place in N-1 steps, each of which shorten the length of the sequence by 1. Let S be a string describing the steps, and the sequence on which the i-th step is performed be (a_1,a_2,...,a_K), then the i-th step will be as follows:

When the i-th character in S is M, let b_i = max(a_i,a_{i+1}) (1 ≦ i ≦ K-1), and replace the current sequence by (b_1,b_2,...,b_{K-1}).

When the i-th character in S is m, let b_i = min(a_i,a_{i+1}) (1 ≦ i ≦ K-1), and replace the current sequence by (b_1,b_2,...,b_{K-1}).

Takahashi decided the steps S, but he is too tired to carry out the compression. On behalf of him, find the single integer obtained from the compression.

Constraints

2 ≦ N ≦ 10^5

1 ≦ A_i ≦ N

S consists of N-1 characters.

Each character in S is either M or m.

Partial Scores

In the test set worth 400 points, there exists i (1 ≦ i < N-1) such that the first i characters in S are M, and the remaining characters are m, that is, S is in the form M...Mm...m.

In the test set worth 800 points, the odd-number-th characters in S are M, and the even-number-th characters are m, that is, S is in the form MmMmMm....

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N
S

Output

Print the single integer obtained from the compression.

Sample Input 1

4
1 2 3 4
MmM

Sample Output 1

3

The initial sequence ( 1 , 2 , 3 , 4 ) is compressed as follows:

After the first step, the sequence is ( 2 , 3 , 4 ).

After the second step, the sequence is ( 2 , 3 ).

After the third step, the sequence is ( 3 ).

Thus, the single integer obtained from the compression is 3.

Sample Input 2

5
3 4 2 2 1
MMmm

Sample Output 2

2

Sample Input 3

10
1 8 7 6 8 5 2 2 6 1
MmmmmMMMm

Sample Output 3

5

Sample Input 4

20
12 7 16 8 7 19 8 19 20 11 7 13 20 3 4 11 19 11 15 5
mMMmmmMMMMMMmMmmmMM

Sample Output 4

11

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>

using namespace std;

typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define Fst first
#define Snd second
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)
#define repv(t,it,v) for(vector<t>::iterator it = v.begin() ; it != v.end() ; ++it)

const int SPT_MAX = 100010;
const int SPT_lg = 20;

template <class T> class SPT{
	T x[SPT_MAX + 1][SPT_lg];
	int lg[SPT_MAX + 1];
	
	T eval(T a , T b){
		return max(a,b);
	}
	
public:
	
	SPT(){
		
	}
	
	void dft(int n , T a[]){
		lg[1] = 0;
		for(int i = 2 ; i <= SPT_MAX ; ++i) lg[i] = lg[i / 2] + 1;
		for(int i = 0 ; i < n ; ++i) x[i][0] = a[i];
		for(int j = 0 ; j < SPT_lg - 1 ; ++j){
			for(int i = 0 ; i < n - (1 << j) ; ++i){
				x[i][j + 1] = eval(x[i][j] , x[i + (1 << j)][j]);
			}
		}
	}
	
	T get(int s , int f){
		int z = f - s;
		return eval(x[s][lg[z]] , x[f - (1 << lg[z])][lg[z]]);
	}
};


int N;
int A[100010];
char s[100010];
SPT<int> T;

int main(){
	scanf("%d" , &N);
	repp(i,0,N){
		scanf("%d" , A + i);
	}
	scanf(" %s" , s);
	int x = 0;
	while(s[x] == 'M') ++x;
	T.dft(N,A);
	int ans = N*2;
	repp(i,1,N-x){
		ans = min(ans , T.get(i,i+x));
	}
	printf("%d\n" , ans);
	return 0;
}