#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool setmin(T &a, T b) {
  if (a > b) return a = b, 1;
  return 0;
}
template <class T>
inline bool setmax(T &a, T b) {
  if (a < b) return a = b, 1;
  return 0;
}
template <class T>
istream &operator>>(istream &os, vector<T> &container) {
  for (auto &u : container) os >> u;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &container) {
  for (auto &u : container) os << u << " ";
  return os;
}
template <typename T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return a;
}
long double e = 2.7182818284590452353602874713526624;
long double PI = acos(-1);
long double PIjane = sqrt(2) + sqrt(3);
long double eps = 1e-11;
const long long T = 1 << 20;
const long long INF = 1e9 + 7;
const long long BIG_INF = 1e18 + 5;
const int N = 1e6 + 7;
long long n, k;
long long tab[N];
map<int, int> mapa;
vector<int> liczby;
long long tree[T << 1][2];
void input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> tab[i];
    liczby.push_back(tab[i]);
  }
}
long long query(int kon, int dim) {
  kon += T;
  int pocz = T;
  long long ret = tree[kon][dim];
  while (pocz / 2 != kon / 2) {
    if (pocz % 2 == 0) ret += tree[pocz + 1][dim];
    if (kon % 2 == 1) ret += tree[kon - 1][dim];
    pocz /= 2;
    kon /= 2;
  }
  return ret;
}
void update(int poz, int val, int dim) {
  poz += T;
  while (poz) {
    tree[poz][dim] += val;
    poz /= 2;
  }
}
void compress() {
  sort((liczby).begin(), (liczby).end());
  unique((liczby).begin(), (liczby).end());
  int licznik = 1;
  for (auto &u : liczby) mapa[u] = licznik++;
}
void solve() {
  long long ret = 0;
  for (int i = n - 1; i >= 0; i--) {
    ret += query(mapa[tab[i]] - 1, 1);
    update(mapa[tab[i]], query(mapa[tab[i]] - 1, 0), 1);
    update(mapa[tab[i]], 1, 0);
  }
  cout << ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  input();
  compress();
  solve();
}

### Prompt

Develop a solution in cpp to the problem described below:

The Romans have attacked again. This time they are much more than the Persians but Shapur is ready to defeat them. He says: "A lion is never afraid of a hundred sheep". 

Nevertheless Shapur has to find weaknesses in the Roman army to defeat them. So he gives the army a weakness number.

In Shapur's opinion the weakness of an army is equal to the number of triplets i, j, k such that i < j < k and ai > aj > ak where ax is the power of man standing at position x. The Roman army has one special trait — powers of all the people in it are distinct.

Help Shapur find out how weak the Romans are.

Input

The first line of input contains a single number n (3 ≤ n ≤ 106) — the number of men in Roman army. Next line contains n different positive integers ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 109) — powers of men in the Roman army. 

Output

A single integer number, the weakness of the Roman army. 

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

Input

3
3 2 1


Output

1


Input

3
2 3 1


Output

0


Input

4
10 8 3 1


Output

4


Input

4
1 5 4 3


Output

1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool setmin(T &a, T b) {
  if (a > b) return a = b, 1;
  return 0;
}
template <class T>
inline bool setmax(T &a, T b) {
  if (a < b) return a = b, 1;
  return 0;
}
template <class T>
istream &operator>>(istream &os, vector<T> &container) {
  for (auto &u : container) os >> u;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &container) {
  for (auto &u : container) os << u << " ";
  return os;
}
template <typename T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return a;
}
long double e = 2.7182818284590452353602874713526624;
long double PI = acos(-1);
long double PIjane = sqrt(2) + sqrt(3);
long double eps = 1e-11;
const long long T = 1 << 20;
const long long INF = 1e9 + 7;
const long long BIG_INF = 1e18 + 5;
const int N = 1e6 + 7;
long long n, k;
long long tab[N];
map<int, int> mapa;
vector<int> liczby;
long long tree[T << 1][2];
void input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> tab[i];
    liczby.push_back(tab[i]);
  }
}
long long query(int kon, int dim) {
  kon += T;
  int pocz = T;
  long long ret = tree[kon][dim];
  while (pocz / 2 != kon / 2) {
    if (pocz % 2 == 0) ret += tree[pocz + 1][dim];
    if (kon % 2 == 1) ret += tree[kon - 1][dim];
    pocz /= 2;
    kon /= 2;
  }
  return ret;
}
void update(int poz, int val, int dim) {
  poz += T;
  while (poz) {
    tree[poz][dim] += val;
    poz /= 2;
  }
}
void compress() {
  sort((liczby).begin(), (liczby).end());
  unique((liczby).begin(), (liczby).end());
  int licznik = 1;
  for (auto &u : liczby) mapa[u] = licznik++;
}
void solve() {
  long long ret = 0;
  for (int i = n - 1; i >= 0; i--) {
    ret += query(mapa[tab[i]] - 1, 1);
    update(mapa[tab[i]], query(mapa[tab[i]] - 1, 0), 1);
    update(mapa[tab[i]], 1, 0);
  }
  cout << ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  input();
  compress();
  solve();
}

```