#include <bits/stdc++.h>
using namespace std;
long long a[1000001], temp[1000001], n, res, seg[5 * 1000001],
    LeftGreater[1000001], RightSmaller[1000001];
map<long long, long long> mp;
void Update(int node, int st, int en, int idx, int value) {
  if (st == en)
    seg[node] += value;
  else {
    int mid = (st + en) / 2;
    if (mid >= idx)
      Update(node * 2, st, mid, idx, value);
    else
      Update(node * 2 + 1, mid + 1, en, idx, value);
    seg[node] = seg[2 * node] + seg[2 * node + 1];
  }
}
int Query(int node, int st, int en, int left, int right) {
  if (left > en || right < st || left > right) return 0;
  if (st >= left && en <= right) return seg[node];
  int mid = (st + en) / 2;
  int p1 = Query(2 * node, st, mid, left, right);
  int p2 = Query(2 * node + 1, mid + 1, en, left, right);
  return p1 + p2;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    temp[i] = a[i];
  }
  sort(temp + 1, temp + n + 1);
  for (int i = 1; i <= n; i++) mp[temp[i]] = i;
  for (int i = 1; i <= n; i++) a[i] = mp[a[i]];
  for (int i = 1; i <= n; i++) {
    LeftGreater[i] = Query(1, 1, n, a[i] + 1, n);
    Update(1, 1, n, a[i], 1);
  }
  memset(seg, 0, sizeof seg);
  for (long long i = n; i >= 1; i--) {
    RightSmaller[i] = Query(1, 1, n, 1, a[i] - 1);
    Update(1, 1, n, a[i], 1);
  }
  for (int i = 1; i <= n; i++) res += RightSmaller[i] * LeftGreater[i];
  cout << res << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 a[1000001], temp[1000001], n, res, seg[5 * 1000001],
    LeftGreater[1000001], RightSmaller[1000001];
map<long long, long long> mp;
void Update(int node, int st, int en, int idx, int value) {
  if (st == en)
    seg[node] += value;
  else {
    int mid = (st + en) / 2;
    if (mid >= idx)
      Update(node * 2, st, mid, idx, value);
    else
      Update(node * 2 + 1, mid + 1, en, idx, value);
    seg[node] = seg[2 * node] + seg[2 * node + 1];
  }
}
int Query(int node, int st, int en, int left, int right) {
  if (left > en || right < st || left > right) return 0;
  if (st >= left && en <= right) return seg[node];
  int mid = (st + en) / 2;
  int p1 = Query(2 * node, st, mid, left, right);
  int p2 = Query(2 * node + 1, mid + 1, en, left, right);
  return p1 + p2;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    temp[i] = a[i];
  }
  sort(temp + 1, temp + n + 1);
  for (int i = 1; i <= n; i++) mp[temp[i]] = i;
  for (int i = 1; i <= n; i++) a[i] = mp[a[i]];
  for (int i = 1; i <= n; i++) {
    LeftGreater[i] = Query(1, 1, n, a[i] + 1, n);
    Update(1, 1, n, a[i], 1);
  }
  memset(seg, 0, sizeof seg);
  for (long long i = n; i >= 1; i--) {
    RightSmaller[i] = Query(1, 1, n, 1, a[i] - 1);
    Update(1, 1, n, a[i], 1);
  }
  for (int i = 1; i <= n; i++) res += RightSmaller[i] * LeftGreater[i];
  cout << res << endl;
  return 0;
}

```