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`.

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.

Examples

Input

4
1 2 3 4
MmM


Output

3


Input

5
3 4 2 2 1
MMmm


Output

2


Input

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


Output

5


Input

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


Output

11
#include <bits/stdc++.h>

using namespace std;

typedef unsigned __int128 TYPE;
const int BASE=128;
TYPE HIGH=((TYPE)1)<<(BASE-1);
int N;
int A[100000];
TYPE B[(100000+BASE-1)/BASE+3];
char S[100000];

void orb(int n)
{
    for(int i=0; i*BASE<n; i++)
        B[i]|=(B[i]>>1)|((B[i+1]&1)<<(BASE-1));
}

void andb(int n)
{
    for(int i=0; i*BASE<n; i++)
        B[i]&=(B[i]>>1)|((B[i+1]&1)<<(BASE-1));
}

bool solve()
{
    for(int i=0; i<N-1; i++)
    {
        if(S[i]=='M')
            orb(N-i);
        else
            andb(N-i);
    }
    return B[0]&1;
}

int main()
{
    scanf("%d", &N);
    for(int i=0; i<N; i++)
        scanf("%d", A+i);
    scanf("%s", S);
    int lo=1, mid, hi=N;
    while(lo<hi)
    {
        mid=(lo+hi+1)/2;
        for(int i=0; i*BASE<N; i++)
            B[i]=0;
        for(int i=0; i<N; i++)
            B[i/BASE]|=((TYPE)(A[i]>=mid))<<(i%BASE);
        if(solve())
            lo=mid;
        else
            hi=mid-1;
    }
    printf("%d\n", lo);
    return 0;
}
