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 <bits/stdc++.h>
using namespace std;
//#include <boost/multiprecision/cpp_int.hpp>
//using multiInt = boost::multiprecision::cpp_int;

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename Q_temp>
using smaller_queue = priority_queue <Q_temp, vector<Q_temp>, greater<Q_temp> >;

const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
const ll MOD = (ll) (1e9 + 7);
const double PI = acos(-1.0);
const int limit = 200010;

#define REP(i,m,n) for(ll i = m; i < (ll)(n); ++i)
#define rep(i,n) REP(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbg2(x_) for(auto a_ : x_) cerr << a_ << " "; cerr << endl;
#define dbg3(x_ , sx_) rep(i, sx_) cerr << x_[i] << " "; cerr << endl;
vector<int> Dx = {0, 0, -1, 1, -1, 1, -1, 1, 0};
vector<int> Dy = {1, -1, 0, 0, -1, -1, 1, 1, 0};

//------------------------------------------------------

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    auto M = [&](char p1, char p2) {
        return (p1 == 'T' && p2 == 'F' ? 'F' : 'T');
    };
    int n;
    cin >> n;
    char p[n];
    rep(i, n) cin >> p[i];
    char now = M(p[0], p[1]);
    for (int i = 2; i < n; ++i) {
        now = M(now, p[i]);
    }
    cout << now << endl;
    return 0;
}
