#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n;
long long ans;
int pref[N];
long long st[4 * N];
pair<int, int> ar[N];
map<int, int> pos;
bool cmp(pair<int, int> a, pair<int, int> b) { return (a.first > b.first); }
void build(int idx, int l, int r) {
  if (l == r) {
    st[idx] = 0;
    return;
  }
  int m = (l + r) / 2;
  build(2 * idx, l, m);
  build(2 * idx + 1, m + 1, r);
  st[idx] = 0;
}
void upd(int idx, int l, int r, int target, int val) {
  if (l == r) {
    st[idx] = val;
    return;
  }
  int m = (l + r) / 2;
  if (target <= m)
    upd(2 * idx, l, m, target, val);
  else
    upd(2 * idx + 1, m + 1, r, target, val);
  st[idx] = st[2 * idx] + st[2 * idx + 1];
}
long long getSum(int idx, int l, int r, int x, int y) {
  if ((l >= x) && (r <= y)) return st[idx];
  if ((y < l) || (x > r)) return 0;
  int m = (l + r) / 2;
  return getSum(2 * idx, l, m, x, y) + getSum(2 * idx + 1, m + 1, r, x, y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i].first;
    ar[i].second = i;
  }
  sort(ar + 1, ar + n + 1, cmp);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    pref[ar[i].second] = getSum(1, 1, n, 1, ar[i].second);
    upd(1, 1, n, ar[i].second, 1);
  }
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    ans += getSum(1, 1, n, 1, ar[i].second);
    upd(1, 1, n, ar[i].second, pref[ar[i].second]);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please formulate 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;
const int N = 1e6 + 5;
int n;
long long ans;
int pref[N];
long long st[4 * N];
pair<int, int> ar[N];
map<int, int> pos;
bool cmp(pair<int, int> a, pair<int, int> b) { return (a.first > b.first); }
void build(int idx, int l, int r) {
  if (l == r) {
    st[idx] = 0;
    return;
  }
  int m = (l + r) / 2;
  build(2 * idx, l, m);
  build(2 * idx + 1, m + 1, r);
  st[idx] = 0;
}
void upd(int idx, int l, int r, int target, int val) {
  if (l == r) {
    st[idx] = val;
    return;
  }
  int m = (l + r) / 2;
  if (target <= m)
    upd(2 * idx, l, m, target, val);
  else
    upd(2 * idx + 1, m + 1, r, target, val);
  st[idx] = st[2 * idx] + st[2 * idx + 1];
}
long long getSum(int idx, int l, int r, int x, int y) {
  if ((l >= x) && (r <= y)) return st[idx];
  if ((y < l) || (x > r)) return 0;
  int m = (l + r) / 2;
  return getSum(2 * idx, l, m, x, y) + getSum(2 * idx + 1, m + 1, r, x, y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i].first;
    ar[i].second = i;
  }
  sort(ar + 1, ar + n + 1, cmp);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    pref[ar[i].second] = getSum(1, 1, n, 1, ar[i].second);
    upd(1, 1, n, ar[i].second, 1);
  }
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    ans += getSum(1, 1, n, 1, ar[i].second);
    upd(1, 1, n, ar[i].second, pref[ar[i].second]);
  }
  cout << ans << '\n';
  return 0;
}

```