#include <bits/stdc++.h>
using namespace std;
long long lst, fen[10 * 100500], a[10 * 100500], c[10 * 100500], l[10 * 100500],
    n;
long long ans;
bool cmp(int x, int y) { return a[x] < a[y]; }
void upd(int x, int val) {
  while (x < n) {
    fen[x] += val;
    x = (x | (x + 1));
  }
}
long long f(int x) {
  long long k = 0;
  while (x >= 0) {
    k += fen[x];
    x = (x & (x + 1)) - 1;
  }
  return k;
}
long long sum(int l, int r) { return f(r) - f(l - 1); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], c[i] = i;
  sort(c, c + n, cmp);
  lst = 0;
  for (int i = 0; i < n; i++) a[c[i]] = lst++;
  for (int i = 0; i < n; i++) {
    l[i] = sum(a[i] + 1, n - 1);
    upd(a[i], 1);
  }
  for (int i = 0; i < n; i++) fen[i] = 0;
  for (int i = n - 1; i >= 0; i--) {
    ans += l[i] * sum(0, a[i]);
    upd(a[i], 1);
  }
  cout << ans;
}

### Prompt

Your task is to create 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;
long long lst, fen[10 * 100500], a[10 * 100500], c[10 * 100500], l[10 * 100500],
    n;
long long ans;
bool cmp(int x, int y) { return a[x] < a[y]; }
void upd(int x, int val) {
  while (x < n) {
    fen[x] += val;
    x = (x | (x + 1));
  }
}
long long f(int x) {
  long long k = 0;
  while (x >= 0) {
    k += fen[x];
    x = (x & (x + 1)) - 1;
  }
  return k;
}
long long sum(int l, int r) { return f(r) - f(l - 1); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], c[i] = i;
  sort(c, c + n, cmp);
  lst = 0;
  for (int i = 0; i < n; i++) a[c[i]] = lst++;
  for (int i = 0; i < n; i++) {
    l[i] = sum(a[i] + 1, n - 1);
    upd(a[i], 1);
  }
  for (int i = 0; i < n; i++) fen[i] = 0;
  for (int i = n - 1; i >= 0; i--) {
    ans += l[i] * sum(0, a[i]);
    upd(a[i], 1);
  }
  cout << ans;
}

```