#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 7;
long long int int_pow(long long int base, long long int exp) {
  long long int result = 1;
  while (exp) {
    if (exp & 1) result *= base;
    exp /= 2;
    base *= base;
  }
  return result;
}
long long int FEN[MAXN], a[MAXN], b[MAXN], r[MAXN], n;
void Update(long long int x) {
  for (; x <= n; x += (x & (-x))) FEN[x]++;
}
long long int Sum(long long int x) {
  long long int ans = 0;
  for (; x > 0; x -= (x & (-x))) ans += FEN[x];
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) cin >> b[i], a[i] = b[i];
  sort(b, b + n);
  for (long long int i = 0; i < n; i++) {
    a[i] = upper_bound(b, b + n, a[i]) - b;
    r[i] = i - Sum(a[i]);
    Update(a[i]);
  }
  fill_n(FEN, MAXN, 0);
  for (long long int i = n - 1; i >= 0; i--) {
    ans += r[i] * Sum(a[i] - 1);
    Update(a[i]);
  }
  cout << ans;
}

### Prompt

Your challenge is to write 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 + 7;
long long int int_pow(long long int base, long long int exp) {
  long long int result = 1;
  while (exp) {
    if (exp & 1) result *= base;
    exp /= 2;
    base *= base;
  }
  return result;
}
long long int FEN[MAXN], a[MAXN], b[MAXN], r[MAXN], n;
void Update(long long int x) {
  for (; x <= n; x += (x & (-x))) FEN[x]++;
}
long long int Sum(long long int x) {
  long long int ans = 0;
  for (; x > 0; x -= (x & (-x))) ans += FEN[x];
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) cin >> b[i], a[i] = b[i];
  sort(b, b + n);
  for (long long int i = 0; i < n; i++) {
    a[i] = upper_bound(b, b + n, a[i]) - b;
    r[i] = i - Sum(a[i]);
    Update(a[i]);
  }
  fill_n(FEN, MAXN, 0);
  for (long long int i = n - 1; i >= 0; i--) {
    ans += r[i] * Sum(a[i] - 1);
    Update(a[i]);
  }
  cout << ans;
}

```