#include <bits/stdc++.h>
using namespace std;
int n, a[1000100], b[1000100];
;
long long one[1000100], two[1000100], ans;
long long sum(long long* a, int i) {
  long long res = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) res += a[i];
  return res;
}
void inc(long long* a, int i, long long delta) {
  for (; i < n; i = i | (i + 1)) a[i] += delta;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    int l = 0, r = n;
    while (r - l > 1) {
      int m = (l + r) >> 1;
      if (b[m] <= a[i])
        l = m;
      else
        r = m;
    }
    a[i] = l;
  }
  reverse(a, a + n);
  for (int i = 0; i < n; i++) {
    inc(one, a[i], 1);
    inc(two, a[i], sum(one, a[i] - 1));
    ans += sum(two, a[i] - 1);
  }
  cout << ans;
  return 0;
}

### 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;
int n, a[1000100], b[1000100];
;
long long one[1000100], two[1000100], ans;
long long sum(long long* a, int i) {
  long long res = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) res += a[i];
  return res;
}
void inc(long long* a, int i, long long delta) {
  for (; i < n; i = i | (i + 1)) a[i] += delta;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    int l = 0, r = n;
    while (r - l > 1) {
      int m = (l + r) >> 1;
      if (b[m] <= a[i])
        l = m;
      else
        r = m;
    }
    a[i] = l;
  }
  reverse(a, a + n);
  for (int i = 0; i < n; i++) {
    inc(one, a[i], 1);
    inc(two, a[i], sum(one, a[i] - 1));
    ans += sum(two, a[i] - 1);
  }
  cout << ans;
  return 0;
}

```