1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
#include <bits/stdc++.h>
using namespace std;
#define IOS                       \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define repV(i, a, n) for (int i = n; i >= a; i--)
#define st first
#define nd second
#define pb push_back
#define mp make_pair
#define all(a) a.begin(), a.end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
constexpr int M = 5e5 + 5;
pair<ll, ll> t[M];
ll akt;
int n, idx;
string ans;

bool check(int x)
{
    akt = t[x].st;
    rep(i, 1, n)
    {
        if (i == x)
            continue;
        if (t[i].st < akt)
            akt += t[i].st;
        else
            return false;
    }
    return true;
}

int bs(int p, int k)
{
    while (p < k)
    {
        int m = (p + k) / 2;
        if (!check(m))
        {
            p = m + 1;
        }
        else
            k = m;
    }
    return p;
}

void solve()
{
    cin >> n;
    rep(i, 1, n)
    {
        cin >> t[i].st;
        t[i].nd = i - 1;
    }
    rep(i, 0, n - 1)
        ans += 'N';
    sort(t + 1, t + 1 + n);
    idx = bs(1, n);
    rep(i, idx, n)
    {
        ans[t[i].nd] = 'T';
    }
    if (!check(n))
        ans[t[n].nd] = 'N';
    else
        ans[t[n].nd] = 'T';
    cout << ans << endl;
}

int main()
{
    IOS
    solve();
}