#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
const int n_ = 1e6 + 5;
int n, arr[n_], b[n_], a[n_], st[n_ << 2], x;
vector<int> v;
map<int, int> m;
long long ans;
void upd(int ind, int id = 1, int l = 1, int r = n) {
  if (l == r) {
    ++st[id];
    return;
  }
  int mid = (l + r) >> 1;
  if (ind > mid)
    upd(ind, id << 1 | 1, mid + 1, r);
  else
    upd(ind, id << 1, l, mid);
  st[id] = st[id << 1] + st[id << 1 | 1];
}
int query(int s, int e, int id = 1, int l = 1, int r = n) {
  if (s <= l && r <= e) return st[id];
  if (s > r || l > e) return 0;
  int mid = (l + r) >> 1;
  return query(s, e, id << 1, l, mid) + query(s, e, id << 1 | 1, mid + 1, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    v.push_back(arr[i]);
  }
  sort(v.begin(), v.end());
  for (int i : v) m[i] = ++x;
  for (int i = 0; i < n; i++) arr[i] = m[arr[i]];
  for (int i = 0; i < n; i++) {
    upd(arr[i]);
    b[i] = query(arr[i] + 1, n);
  }
  memset(st, 0, sizeof st);
  for (int i = n - 1; ~i; --i) {
    upd(arr[i]);
    a[i] = query(1, arr[i] - 1);
  }
  for (int i = 0; i < n; i++) ans += (long long)a[i] * b[i];
  printf("%lld", ans);
  return 0;
}

### Prompt

Generate 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 double pi = acos(-1);
const int n_ = 1e6 + 5;
int n, arr[n_], b[n_], a[n_], st[n_ << 2], x;
vector<int> v;
map<int, int> m;
long long ans;
void upd(int ind, int id = 1, int l = 1, int r = n) {
  if (l == r) {
    ++st[id];
    return;
  }
  int mid = (l + r) >> 1;
  if (ind > mid)
    upd(ind, id << 1 | 1, mid + 1, r);
  else
    upd(ind, id << 1, l, mid);
  st[id] = st[id << 1] + st[id << 1 | 1];
}
int query(int s, int e, int id = 1, int l = 1, int r = n) {
  if (s <= l && r <= e) return st[id];
  if (s > r || l > e) return 0;
  int mid = (l + r) >> 1;
  return query(s, e, id << 1, l, mid) + query(s, e, id << 1 | 1, mid + 1, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    v.push_back(arr[i]);
  }
  sort(v.begin(), v.end());
  for (int i : v) m[i] = ++x;
  for (int i = 0; i < n; i++) arr[i] = m[arr[i]];
  for (int i = 0; i < n; i++) {
    upd(arr[i]);
    b[i] = query(arr[i] + 1, n);
  }
  memset(st, 0, sizeof st);
  for (int i = n - 1; ~i; --i) {
    upd(arr[i]);
    a[i] = query(1, arr[i] - 1);
  }
  for (int i = 0; i < n; i++) ans += (long long)a[i] * b[i];
  printf("%lld", ans);
  return 0;
}

```