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// {{{
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define FOR(i, b, e) for (int i = (b); i <= (e); i++)
#define FORD(i, b, e) for (int i = (e); i >= (b); i--)
#define MP make_pair
#define FS first
#define ND second
#define PB push_back
#define ALL(x) x.begin(), x.end()

using namespace std;

using LL = long long;
using LD = long double;
using PII = pair<int,int>;
using PLL = pair<LL,LL>;
using PLD = pair<LD,LD>;
using VI = vector<int>;
using VLL = vector<LL>;
using VLD = vector<LD>;
using VB = vector<bool>;
using VS = vector<string>;
template<class T>
using V = vector<T>;
template<class T1, class T2>
using P = pair<T1,T2>;
template<class T,class Comp=greater<T>>
using PQ = priority_queue<T,V<T>,Comp>;

template<class T>
int sz(const T& a) { return (int)a.size(); }
template<class T>
void amin(T& a, T b) { a = min(a, b); }
template<class T>
void amax(T& a, T b) { a = max(a, b); }

/*
const size_t rseed =
  std::chrono::high_resolution_clock::now().time_since_epoch().count();
mt19937 rnd(rseed);
template<class T> T randint(T lo, T hi)
{
  return uniform_int_distribution<T>{lo,hi}(rnd);
}
*/

/* find_by_order(k) - k'th largest counting from 0;
   order_of_key(k) - number of items strictly smaller than k;
   join(other), split(k, other) - all keys in other greater than in *(this). */
template<class K,class V,class Comp=less<K>>
using ordered_map = __gnu_pbds::tree<K,V,Comp,
      __gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>;
template<class T,class Comp=less<T>>
using ordered_set = ordered_map<T,__gnu_pbds::null_type,Comp>;

const int inf = 1e9 + 1;
const LL infl = 1e18 + 1;

void solve();
// }}}

int main()
{
  int t = 1;
  // scanf("%d", &t);
  FOR(i, 1, t) {
    // printf("Case #%d: ", i);
    solve();
  }
}

const int N = 1e6;

int n, p[N + 1];
LL ans = 0;

// ile sposob mozna zapisac c = x + y, gdzie x <= a, y <= b
LL f(int a, int b, int c)
{
  return max(0, min(a, c) - max(0, c - b) + 1);
}

void solve()
{
  scanf("%d", &n);
  FOR(i, 1, n) {
    int x;
    scanf("%d", &x);
    p[x] = i;
  }

  int l = p[n], r = p[n];
  FORD(x, 1, n) {
    while (p[x] < l) l--;
    while (r < p[x]) r++;

    // odd length with median x
    ans += f(l - 1, n - r, 2 * (n - x) + 1 - (r - l + 1));
    // even length with median x + 0.5
    ans += f(l - 1, n - r, 2 * (n - x) - (r - l + 1));
  }

  printf("%d %lld\n", 2 * n + 1, ans);
}