#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using dd = double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
ll INFLL = (ll)2e18, MOD = 1e9 + 7;
const int INF = 0x6f6f6f6f;
vector<vector<ll>> adj;
vector<int> vis;
int dx8[] = {0, 1, 1, 1, 0, -1, -1, -1}, dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
    dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
inline ll mexp(ll x, ll n, ll m) {
  ll res = 1;
  while (n) {
    if (n & 1) res = (res * x) % m;
    n >>= 1;
    x = ((x % m) * (x % m)) % m;
  }
  return res;
}
inline bool ispow2(ll x) { return x && (!(x & (x - 1))); }
inline ll gcd(ll x, ll y) {
  pll p{x, y};
  while (p.second) p = {p.second, p.first % p.second};
  return p.first;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed;
  cerr << fixed;
  cout << setprecision(10);
  cerr << setprecision(3);
  mt19937 genr(chrono::high_resolution_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  vector<int> seq(n + 1);
  map<int, int> m;
  for (int i = 0; i < n; ++i) {
    int t;
    cin >> t;
    m[t] = i + 1;
  }
  int ct = 1;
  for (auto& e : m) seq[e.second] = ct++;
  m.clear();
  vector<ll> tree(n + 1, 0);
  auto sum = [&](int pos) {
    ll ret = 0;
    while (pos) {
      ret += tree[pos];
      pos -= pos & -pos;
    }
    return ret;
  };
  auto upd = [&](int pos, ll val) {
    while (pos <= n) {
      tree[pos] += val;
      pos += pos & -pos;
    }
  };
  ll ans = 0;
  for (int i = 1; i <= n; ++i) {
    ll gr = i - 1 - sum(seq[i]);
    ll le = seq[i] - 1 - (i - 1 - gr);
    ans += gr * le;
    upd(seq[i], 1);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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;
using ll = long long int;
using dd = double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
ll INFLL = (ll)2e18, MOD = 1e9 + 7;
const int INF = 0x6f6f6f6f;
vector<vector<ll>> adj;
vector<int> vis;
int dx8[] = {0, 1, 1, 1, 0, -1, -1, -1}, dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
    dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
inline ll mexp(ll x, ll n, ll m) {
  ll res = 1;
  while (n) {
    if (n & 1) res = (res * x) % m;
    n >>= 1;
    x = ((x % m) * (x % m)) % m;
  }
  return res;
}
inline bool ispow2(ll x) { return x && (!(x & (x - 1))); }
inline ll gcd(ll x, ll y) {
  pll p{x, y};
  while (p.second) p = {p.second, p.first % p.second};
  return p.first;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed;
  cerr << fixed;
  cout << setprecision(10);
  cerr << setprecision(3);
  mt19937 genr(chrono::high_resolution_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  vector<int> seq(n + 1);
  map<int, int> m;
  for (int i = 0; i < n; ++i) {
    int t;
    cin >> t;
    m[t] = i + 1;
  }
  int ct = 1;
  for (auto& e : m) seq[e.second] = ct++;
  m.clear();
  vector<ll> tree(n + 1, 0);
  auto sum = [&](int pos) {
    ll ret = 0;
    while (pos) {
      ret += tree[pos];
      pos -= pos & -pos;
    }
    return ret;
  };
  auto upd = [&](int pos, ll val) {
    while (pos <= n) {
      tree[pos] += val;
      pos += pos & -pos;
    }
  };
  ll ans = 0;
  for (int i = 1; i <= n; ++i) {
    ll gr = i - 1 - sum(seq[i]);
    ll le = seq[i] - 1 - (i - 1 - gr);
    ans += gr * le;
    upd(seq[i], 1);
  }
  cout << ans << '\n';
  return 0;
}

```