#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }
#define COLOR(s) ("\x1b[" s "m")


const double L = 200;

int N;
vector<int> P;

int main() {
  for (; ~scanf("%d", &N); ) {
    P.assign(N + 2, 0);
    for (int i = 1; i <= N; ++i) {
      scanf("%d", &P[i]);
    }
    
    Int ans = 0;
Int sumM=0;
    vector<int> freq(2 * N + 3, 0);
    for (int i = 1; i <= N; ++i) {
      /*
        interval of length m + 1
        binomial distribution
          Pr[small] = p
          E[# small] = m p
          V[# small] = m p (1 - p)
        E[large - small] = m (1 - 2 p)
        V[large - small] = 4 m p (1 - p)
        let 0 or +1 lie within m (1 - 2 p) +- L * 2 sqrt(m p (1 - p))
      */
      const int a = P[i];
      const double p = (a - 1) / (double)(N - 1);
      int m = L * 2 * sqrt(p * (1 - p)) / max(abs(1 - 2 * p), 1e-6);
      m = min(max(m, 100), N);
sumM+=m;
      fill(freq.begin(), freq.begin() + (2 * m + 3), 0);
      int now = m + 1;
      ++freq[now];
      for (int j = i - 1; j >= 1 && j >= i - m; --j) {
        (P[j] < a) ? --now : ++now;
        ++freq[now];
      }
      Int here = 0;
      now = m + 1;
      here += freq[2 * m + 2 - now] + freq[2 * m + 3 - now];
      for (int j = i + 1; j <= N && j <= i + m; ++j) {
        (P[j] < a) ? --now : ++now;
        here += freq[2 * m + 2 - now] + freq[2 * m + 3 - now];
      }
// cerr<<a<<": "<<m<<" "<<here<<endl;
      ans += here * a;
    }
cerr<<"sumM = "<<sumM<<endl;
    printf("%lld\n", ans);
  }
  return 0;
}