#include <bits/stdc++.h>
using namespace std;
long long n, ind, ans = 0;
vector<long long> a, b, ft, inv;
void upd(long long x) {
  for (long long i = x; i < ft.size(); i += (i & -i)) {
    ft[i]++;
  }
}
void upd(long long x, long long val) {
  for (long long i = x; i < ft.size(); i += (i & -i)) {
    ft[i] += val;
  }
}
long long calc(long long x) {
  long long ret = 0;
  while (x) {
    ret += ft[x];
    x -= (x & -x);
  }
  return ret;
}
void solve() {
  cin >> n;
  a.resize(n), b.resize(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b.begin(), b.end());
  for (long long i = 0; i < n; ++i) {
    ind = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
    ind++;
    a[i] = ind;
  }
  ft = vector<long long>(n + 1, 0);
  inv = vector<long long>(n, 0);
  for (long long i = n - 1; i >= 0; --i) {
    inv[i] = calc(a[i]);
    upd(a[i]);
  }
  ft = vector<long long>(n + 1, 0);
  for (long long i = n - 1; i >= 0; --i) {
    ans += calc(a[i]);
    upd(a[i], inv[i]);
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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;
long long n, ind, ans = 0;
vector<long long> a, b, ft, inv;
void upd(long long x) {
  for (long long i = x; i < ft.size(); i += (i & -i)) {
    ft[i]++;
  }
}
void upd(long long x, long long val) {
  for (long long i = x; i < ft.size(); i += (i & -i)) {
    ft[i] += val;
  }
}
long long calc(long long x) {
  long long ret = 0;
  while (x) {
    ret += ft[x];
    x -= (x & -x);
  }
  return ret;
}
void solve() {
  cin >> n;
  a.resize(n), b.resize(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b.begin(), b.end());
  for (long long i = 0; i < n; ++i) {
    ind = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
    ind++;
    a[i] = ind;
  }
  ft = vector<long long>(n + 1, 0);
  inv = vector<long long>(n, 0);
  for (long long i = n - 1; i >= 0; --i) {
    inv[i] = calc(a[i]);
    upd(a[i]);
  }
  ft = vector<long long>(n + 1, 0);
  for (long long i = n - 1; i >= 0; --i) {
    ans += calc(a[i]);
    upd(a[i], inv[i]);
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```