#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
long long seg[2][4 * maxn], n;
pair<int, int> a[maxn];
void add(int idx, int val, int x, int id = 1, int l = 0, int r = maxn) {
  seg[idx][id] += val;
  if (r - l == 1) return;
  int mid = (r + l) / 2;
  if (x < mid)
    add(idx, val, x, id * 2 + 0, l, mid);
  else
    add(idx, val, x, id * 2 + 1, mid, r);
}
long long get(int idx, int st, int en, int l = 0, int r = maxn, int id = 1) {
  if (l == st and r == en) return seg[idx][id];
  int mid = (r + l) / 2;
  long long ret = 0;
  if (mid > st) ret += get(idx, st, min(mid, en), l, mid, id * 2 + 0);
  if (mid < en) ret += get(idx, max(st, mid), en, mid, r, id * 2 + 1);
  return ret;
}
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.second < b.second) return true;
  return false;
}
int main() {
  long long ans = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  a[0].first = 1;
  add(1, 1, 1);
  for (int i = 1; i < n; i++) {
    if (a[i].first == a[i - 1].first)
      a[i].first = a[i - 1].first;
    else
      a[i].first = a[i - 1].first + 1;
    add(1, 1, a[i].first);
  }
  sort(a, a + n, cmp);
  add(1, -1, a[0].first);
  add(0, 1, a[0].first);
  for (int i = 1; i < n; i++) {
    ans += (get(1, 0, a[i].first) * get(0, a[i].first + 1, maxn));
    add(1, -1, a[i].first);
    add(0, 1, a[i].first);
  }
  cout << ans << endl;
  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 int maxn = 1e6 + 10;
long long seg[2][4 * maxn], n;
pair<int, int> a[maxn];
void add(int idx, int val, int x, int id = 1, int l = 0, int r = maxn) {
  seg[idx][id] += val;
  if (r - l == 1) return;
  int mid = (r + l) / 2;
  if (x < mid)
    add(idx, val, x, id * 2 + 0, l, mid);
  else
    add(idx, val, x, id * 2 + 1, mid, r);
}
long long get(int idx, int st, int en, int l = 0, int r = maxn, int id = 1) {
  if (l == st and r == en) return seg[idx][id];
  int mid = (r + l) / 2;
  long long ret = 0;
  if (mid > st) ret += get(idx, st, min(mid, en), l, mid, id * 2 + 0);
  if (mid < en) ret += get(idx, max(st, mid), en, mid, r, id * 2 + 1);
  return ret;
}
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.second < b.second) return true;
  return false;
}
int main() {
  long long ans = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  a[0].first = 1;
  add(1, 1, 1);
  for (int i = 1; i < n; i++) {
    if (a[i].first == a[i - 1].first)
      a[i].first = a[i - 1].first;
    else
      a[i].first = a[i - 1].first + 1;
    add(1, 1, a[i].first);
  }
  sort(a, a + n, cmp);
  add(1, -1, a[0].first);
  add(0, 1, a[0].first);
  for (int i = 1; i < n; i++) {
    ans += (get(1, 0, a[i].first) * get(0, a[i].first + 1, maxn));
    add(1, -1, a[i].first);
    add(0, 1, a[i].first);
  }
  cout << ans << endl;
  return 0;
}

```