problem

There is a mysterious device $ M $, and if you put Tanuki and Fox in this device, one animal will come out from the device (hereinafter, Tanuki will be $ T $ and Fox will be $ F $).

$ M (x, y) $ represents an animal that came out by putting animals in the device $ M $ in the order of $ x, y $.

As a result of various trials, the following was found.


$ M (T, T) = T $
$ M (T, F) = F $
$ M (F, T) = T $
$ M (F, F) = T $



You put the animals in a row $ P_1, P_2, ..., P_N $ into the device as follows.
$ M (.... M (M (P_1, P_2), P_3) ...., P_N) $

Answer the last animal that appears.



output

Output the last animal character $ T $ or $ F $, and also a newline at the end.

Example

Input

3
F T T


Output

T
#include <iostream>
#include <bitset>
#include <fstream>
#include <string>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <vector>
#include <algorithm>
#include <numeric>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <functional>
#include <cctype>
#include <list>
#include <limits>
//#include <boost/multiprecision/cpp_int.hpp>

const double EPS = (1e-10);


using namespace std;
using Int = long long;
//using namespace boost::multiprecision;

const Int MOD = 1000000007;

const int MAXN = 1000000;
Int fact[MAXN], inv[MAXN];

Int mod_pow(Int x, Int n) {
    Int res = 1;
    while(n > 0) {
        if(n & 1) res = (res * x) % MOD; //ビット演算(最下位ビットが1のとき)
        x = (x * x) % MOD;
        n >>= 1; //右シフト(n = n >> 1)
    }
    return res;
}

void init_fact() {
    fact[0] = 1;
    for(int i=1; i<MAXN; i++) {
        fact[i] = (fact[i-1] * i) % MOD;
    }

    inv[MAXN - 1] = mod_pow(fact[MAXN - 1], MOD-2);
    for(int i=MAXN - 2; i>=0; i--) {
        inv[i] = (inv[i+1] * (i+1)) % MOD;
    }
}

Int comb(int n, int r) {
    if(r < 0 || n < r) return 0;
    return fact[n] * inv[n-r] % MOD * inv[r] % MOD;
}

template<typename T>
T gcd(T a, T b) {
    return b != 0 ? gcd(b, a % b) : a;
}

template<typename T>
T lcm(T a, T b) {
    return a * b / gcd(a, b);
}

char s[100005];
char st;
void rec(char a, char b){
    if (a == 'T' && b == 'F') st = 'F';
    else st = 'T';
}

int main(){
    cin.tie(0);
    
    int N; cin >> N;
    for (int i = 0; i < N; i++){
        cin >> s[i];
    }
    st = s[0];
    for (int i = 1; i < N; i++){
        rec(st, s[i]);
    }
    cout << st << endl;

}

