#include <bits/stdc++.h>
using ll = long long;
const ll N = 3 * 1e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 1e18;
const ll SZ = 101;
const double eps = 1e-9;
using namespace std;
ll n;
ll BIT[2][10 * N + 1];
void add(ll t, ll pos, ll val) {
  while (pos <= N * 10) {
    BIT[t][pos] += val;
    pos += pos & -pos;
  }
}
ll pfx(ll pos, ll t) {
  ll res = 0;
  while (pos) {
    res += BIT[t][pos];
    pos -= pos & -pos;
  }
  return res;
}
void solve() {
  cin >> n;
  ll ans = 0;
  std::vector<ll> arr(n);
  vector<pair<ll, ll> > a(n);
  for (ll(i) = 0; i < (n); i++)
    cin >> arr[i], a[i].first = arr[i], a[i].second = i;
  sort((a).begin(), (a).end());
  for (ll(i) = 0; i < (n); i++) arr[a[i].second] = i + 1;
  for (ll(i) = 0; i < (n); i++) {
    ll x = arr[i];
    ll tmp = pfx(10 * N, 0) - pfx(x, 0);
    add(0, x, 1);
    ans += pfx(10 * N, 1) - pfx(x, 1);
    add(1, x, tmp);
  }
  cout << ans;
}
int32_t main(int32_t argc, char *argv[]) {
  double t1 = clock();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll TC = 1, t = 0;
  while (t++ < TC) {
    solve();
  }
  if (false)
    cerr << "Time : " << 1000 * (clock() - t1) / CLOCKS_PER_SEC << " ms\n";
  return 0;
}

### Prompt

Create a solution in Cpp for 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 ll = long long;
const ll N = 3 * 1e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 1e18;
const ll SZ = 101;
const double eps = 1e-9;
using namespace std;
ll n;
ll BIT[2][10 * N + 1];
void add(ll t, ll pos, ll val) {
  while (pos <= N * 10) {
    BIT[t][pos] += val;
    pos += pos & -pos;
  }
}
ll pfx(ll pos, ll t) {
  ll res = 0;
  while (pos) {
    res += BIT[t][pos];
    pos -= pos & -pos;
  }
  return res;
}
void solve() {
  cin >> n;
  ll ans = 0;
  std::vector<ll> arr(n);
  vector<pair<ll, ll> > a(n);
  for (ll(i) = 0; i < (n); i++)
    cin >> arr[i], a[i].first = arr[i], a[i].second = i;
  sort((a).begin(), (a).end());
  for (ll(i) = 0; i < (n); i++) arr[a[i].second] = i + 1;
  for (ll(i) = 0; i < (n); i++) {
    ll x = arr[i];
    ll tmp = pfx(10 * N, 0) - pfx(x, 0);
    add(0, x, 1);
    ans += pfx(10 * N, 1) - pfx(x, 1);
    add(1, x, tmp);
  }
  cout << ans;
}
int32_t main(int32_t argc, char *argv[]) {
  double t1 = clock();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll TC = 1, t = 0;
  while (t++ < TC) {
    solve();
  }
  if (false)
    cerr << "Time : " << 1000 * (clock() - t1) / CLOCKS_PER_SEC << " ms\n";
  return 0;
}

```