#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e6 + 6;
int ara[MAX], bit[MAX];
long long bame[MAX], dane[MAX];
int n;
map<int, int> id;
void add(int idx, int delta) {
  while (idx < n) {
    bit[idx] += delta;
    idx = idx | (idx + 1);
  }
}
int sum(int R) {
  int ret = 0;
  while (R >= 0) {
    ret += bit[R];
    R = (R & (R + 1)) - 1;
  }
  return ret;
}
int rangeSum(int L, int R) {
  if (L > R) return 0;
  return sum(R) - sum(L - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ara[i];
    id[ara[i]] = i;
  }
  sort(ara, ara + n);
  for (int i = 0; i < n; i++) {
    add(id[ara[i]], 1);
    dane[id[ara[i]]] = rangeSum(id[ara[i]] + 1, n - 1);
  }
  reverse(ara, ara + n);
  memset(bit, 0, sizeof bit);
  for (int i = 0; i < n; i++) {
    add(id[ara[i]], 1);
    bame[id[ara[i]]] = rangeSum(0, id[ara[i]] - 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += bame[i] * dane[i];
  }
  cout << ans;
}

### Prompt

Create a solution in cpp for 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 MAX = 1e6 + 6;
int ara[MAX], bit[MAX];
long long bame[MAX], dane[MAX];
int n;
map<int, int> id;
void add(int idx, int delta) {
  while (idx < n) {
    bit[idx] += delta;
    idx = idx | (idx + 1);
  }
}
int sum(int R) {
  int ret = 0;
  while (R >= 0) {
    ret += bit[R];
    R = (R & (R + 1)) - 1;
  }
  return ret;
}
int rangeSum(int L, int R) {
  if (L > R) return 0;
  return sum(R) - sum(L - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ara[i];
    id[ara[i]] = i;
  }
  sort(ara, ara + n);
  for (int i = 0; i < n; i++) {
    add(id[ara[i]], 1);
    dane[id[ara[i]]] = rangeSum(id[ara[i]] + 1, n - 1);
  }
  reverse(ara, ara + n);
  memset(bit, 0, sizeof bit);
  for (int i = 0; i < n; i++) {
    add(id[ara[i]], 1);
    bame[id[ara[i]]] = rangeSum(0, id[ara[i]] - 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += bame[i] * dane[i];
  }
  cout << ans;
}

```