#include <bits/stdc++.h>
using namespace std;
int c[1000005], a[1000005], tmp[1000005];
int pre[1000005], suf[1000005];
int n;
int sum(int i) {
  int s = 0;
  for (; i > 0; i -= i & -i) s += c[i];
  return s;
}
void add(int i) {
  for (; i <= n; i += i & -i) c[i]++;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), tmp[i - 1] = a[i];
  sort(tmp, tmp + n);
  int cnt = unique(tmp, tmp + n) - tmp;
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(tmp, tmp + cnt, a[i]) - tmp + 1;
  for (int i = 1; i <= n; i++) pre[i] = (i - 1 - sum(a[i] - 1)), add(a[i]);
  memset(c, 0, sizeof(c));
  for (int i = n; i >= 1; i--) suf[i] = sum(a[i] - 1), add(a[i]);
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += 1LL * pre[i] * suf[i];
  cout << ans << endl;
  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 c[1000005], a[1000005], tmp[1000005];
int pre[1000005], suf[1000005];
int n;
int sum(int i) {
  int s = 0;
  for (; i > 0; i -= i & -i) s += c[i];
  return s;
}
void add(int i) {
  for (; i <= n; i += i & -i) c[i]++;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), tmp[i - 1] = a[i];
  sort(tmp, tmp + n);
  int cnt = unique(tmp, tmp + n) - tmp;
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(tmp, tmp + cnt, a[i]) - tmp + 1;
  for (int i = 1; i <= n; i++) pre[i] = (i - 1 - sum(a[i] - 1)), add(a[i]);
  memset(c, 0, sizeof(c));
  for (int i = n; i >= 1; i--) suf[i] = sum(a[i] - 1), add(a[i]);
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += 1LL * pre[i] * suf[i];
  cout << ans << endl;
  return 0;
}

```