#include <bits/stdc++.h>
using namespace std;
int n;
long long BIT[1000010];
void update(int x, int val) {
  while (x <= n) {
    BIT[x] += val;
    x += (x & -x);
  }
}
long long sum(int x) {
  long long res = 0;
  while (x) {
    res += BIT[x];
    x -= (x & -x);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  int a[n];
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int b[n];
  for (int i = 0; i < n; i++) b[i] = a[i];
  sort(b, b + n);
  for (int i = 0; i < n; i++) a[i] = 1 + (lower_bound(b, b + n, a[i]) - b);
  long long inv = 0;
  long long prev[n];
  long long next[n];
  for (int i = n - 1; i >= 0; i--) {
    update(a[i], 1);
    prev[i] = sum(a[i] - 1);
  }
  memset(BIT, 0, (n + 1) * sizeof(long long));
  for (int i = 0; i < n; i++) {
    update(a[i], 1);
    next[i] = sum(n) - sum(a[i]);
  }
  for (int i = 1; i < n - 1; i++) inv += prev[i] * next[i];
  cout << inv << endl;
  return 0;
}

### 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;
int n;
long long BIT[1000010];
void update(int x, int val) {
  while (x <= n) {
    BIT[x] += val;
    x += (x & -x);
  }
}
long long sum(int x) {
  long long res = 0;
  while (x) {
    res += BIT[x];
    x -= (x & -x);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  int a[n];
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int b[n];
  for (int i = 0; i < n; i++) b[i] = a[i];
  sort(b, b + n);
  for (int i = 0; i < n; i++) a[i] = 1 + (lower_bound(b, b + n, a[i]) - b);
  long long inv = 0;
  long long prev[n];
  long long next[n];
  for (int i = n - 1; i >= 0; i--) {
    update(a[i], 1);
    prev[i] = sum(a[i] - 1);
  }
  memset(BIT, 0, (n + 1) * sizeof(long long));
  for (int i = 0; i < n; i++) {
    update(a[i], 1);
    next[i] = sum(n) - sum(a[i]);
  }
  for (int i = 1; i < n - 1; i++) inv += prev[i] * next[i];
  cout << inv << endl;
  return 0;
}

```