Problem:
Score : 400 points

Problem Statement

One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.

Find the following:

Constraints

1 ≦ N ≦ 200,000

(a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the answer.

Note that the answer may not fit into a 32-bit integer.

Sample Input 1

3
2 1 3

Sample Output 1

9

Sample Input 2

4
1 3 2 4

Sample Output 2

19

Sample Input 3

8
5 4 8 1 2 6 7 3

Sample Output 3

85

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <random>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef double ld;
#define int ll
//#define double ld
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<vvc> vvvc;
typedef pair<int, int> pii;
typedef pair<pii, pii> piii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector< vpi > vvpi;
typedef vector< vvpi > vvvpi;
typedef vector< vi > vvi;
typedef vector< vvi > vvvi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;

#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < int(n); ++i)
#define SZ(a) ((int)((a).size()))
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cerr << (#x) << " = " << (x) << endl;

#ifdef RUS_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif

const double PI = 2 * acos(0.0);
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";


template <class T0, class T1>
inline ostream & operator << (ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template <class T0, class T1>
inline istream & operator >> (istream &in, pair<T0, T1> &a) {
    return in >> a.first >> a.second;
}

template <class T0, class T1, class T2>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template <class T0, class T1, class T2, class T3>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << ", " <<  get<3>(a) << "}";
}

template<class T>
inline ostream & operator << (ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())
        out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}


void smain();

const int MOD1 = 1000000093, MOD2 = 1000000087, p1 = 27162, p2 = 241;

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
#ifdef RUS_HOME
    freopen("input", "r", stdin);
    clock_t start = clock();
#endif
    cout << setprecision(12) << fixed;
    smain();
#ifdef RUS_HOME
    cout << "\n\n\n\nTOTAL EXECUTION TIME: " << float( clock () - start ) /  CLOCKS_PER_SEC << endl;
#endif
    return 0;
}


void smain() {


    int n;
    cin>>n;
    vi a(n);
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    vi lf(n),rg(n,n-1);
    vi kek;
    for(int i=0;i<n;i++){
        while(SZ(kek)&&a[kek.back()]>a[i])
            kek.pop_back();
        if(SZ(kek))
            lf[i]=kek.back()+1;
        kek.push_back(i);
    }
    kek.clear();
    for(int i=n-1;i>=0;i--){
        while(SZ(kek)&&a[kek.back()]>a[i])
            kek.pop_back();
        if(SZ(kek))
            rg[i]=kek.back()-1;
        kek.push_back(i);
    }
    //watch(lf);
    //watch(rg);

    int ans=0;
    for(int i=0;i<n;i++){
        ans+=a[i]*(rg[i]-i+1)*(i-lf[i]+1);
    }
    cout<<ans;


}