#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[10000000], b[10000000], id[10000000], f[10000000];
int cmp(int x, int y) { return a[x] < a[y]; }
void add(int x, int v) {
  for (int i = x; i <= n; i += i & -i) b[i] += v;
}
int que(int x) {
  int ret = 0;
  for (int i = x; i > 0; i -= i & -i) ret += b[i];
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) scanf("%d", a + i);
  for (int i = (0); i < (n); i++) id[i] = i;
  sort(id, id + n, cmp);
  for (int i = (0); i < (n); i++) f[id[i]] = i + 1;
  long long ans = 0;
  for (int i = (0); i < (n); i++) {
    long long t1, t2;
    t1 = f[i] - 1 - que(f[i]);
    t2 = que(n) - que(f[i]);
    ans += t1 * t2;
    add(f[i], 1);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for 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, m;
int a[10000000], b[10000000], id[10000000], f[10000000];
int cmp(int x, int y) { return a[x] < a[y]; }
void add(int x, int v) {
  for (int i = x; i <= n; i += i & -i) b[i] += v;
}
int que(int x) {
  int ret = 0;
  for (int i = x; i > 0; i -= i & -i) ret += b[i];
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) scanf("%d", a + i);
  for (int i = (0); i < (n); i++) id[i] = i;
  sort(id, id + n, cmp);
  for (int i = (0); i < (n); i++) f[id[i]] = i + 1;
  long long ans = 0;
  for (int i = (0); i < (n); i++) {
    long long t1, t2;
    t1 = f[i] - 1 - que(f[i]);
    t2 = que(n) - que(f[i]);
    ans += t1 * t2;
    add(f[i], 1);
  }
  cout << ans << '\n';
  return 0;
}

```