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
82
83
84
85
86
87
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef pair<ld, ld> PLD;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef vector<VII> VVII;
typedef vector<PLL> VLL;
typedef vector<VLL> VVLL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef vector<bool> VB;
typedef map<int, int> MII;
typedef unordered_map<int, int> UMII;
typedef map<ll, ll> MLL;
typedef unordered_map<ll, ll> UMLL;

template <class T, class G>
ostream &operator<<(ostream &os, const pair<T, G> &para) {
  os << para.first << " " << para.second;
  return os;
}
template <class T> ostream &operator<<(ostream &os, const vector<T> &vec) {
  for (const T &el : vec)
    os << el << " ";
  return os;
}
template <class T> ostream &operator<<(ostream &os, const set<T> &vec) {
  for (const T &el : vec)
    os << el << " ";
  return os;
}

#define REP(i, j) for (int i = 0; i < j; i++)
#define REP1(i, j) for (int i = 1; i < j; i++)
#define FOREACH(el, n) for (auto &el : n)
#define LAST(x) (x[(int)x.size() - 1])
#define pb push_back
#define pf push_front
#define st first
#define nd second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define get_unique(x) x.erase(unique(all(x)), x.end());
#define sp ' '
#define ent '\n'

int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);

  int t = 1;
  // cin >> t;
  while (t--) {
    ll n;
    cin >> n;
    VL v;
    VL ind(n + 5, 0);
    REP(i, n) {
      int a;
      cin >> a;
      v.pb(a);
      ind[a] = i;
    }
    ll l = 1e9;
    ll r = -1;
    ll ans = 0;
    for (ll len = 1; len <= n; len++) {
      ll needed_num = n - ((len + 2) / 2 - 1);
      l = min(ind[needed_num], l);
      r = max(ind[needed_num], r);
      ans += max(0LL, min({len - (r - l + 1) + 1, n - len + 1, n - r, l + 1}));
    }
    cout << 2 * n + 1 << sp << ans;
  }

  return 0;
}