#include <bits/stdc++.h>
using namespace std;
struct Node {
  long long Val, Num;
};
bool operator<(Node a, Node b) {
  if (a.Val != b.Val) return a.Val < b.Val;
  return a.Num < b.Num;
}
long long n;
Node List[1001000];
long long a[1001000], b[1001000];
long long tree[1001000];
void Input() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &List[i].Val);
    List[i].Num = i + 1;
  }
  sort(List, List + n);
}
long long MaxVal = 1001000;
long long read(long long idx) {
  long long sum = 0;
  while (idx > 0) {
    sum += tree[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void update(long long idx, long long val) {
  while (idx <= MaxVal) {
    tree[idx] += val;
    idx += (idx & -idx);
  }
}
void SetA() {
  for (long long i = 0; i < n; i++) {
    a[i] = read(n - List[i].Num + 1);
    update(n - List[i].Num + 1, 1);
  }
}
void SetB() {
  for (long long i = n - 1; i >= 0; i--) {
    b[i] = read(List[i].Num);
    update(List[i].Num, 1);
  }
}
void Output() {
  long long t = 0;
  for (long long i = 0; i < n; i++) {
    t += a[i] * b[i];
  }
  cout << t << endl;
}
int main() {
  Input();
  SetA();
  for (long long i = 0; i < 1001000; i++) tree[i] = 0;
  SetB();
  Output();
  return 0;
}

### Prompt

Please formulate 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;
struct Node {
  long long Val, Num;
};
bool operator<(Node a, Node b) {
  if (a.Val != b.Val) return a.Val < b.Val;
  return a.Num < b.Num;
}
long long n;
Node List[1001000];
long long a[1001000], b[1001000];
long long tree[1001000];
void Input() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &List[i].Val);
    List[i].Num = i + 1;
  }
  sort(List, List + n);
}
long long MaxVal = 1001000;
long long read(long long idx) {
  long long sum = 0;
  while (idx > 0) {
    sum += tree[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void update(long long idx, long long val) {
  while (idx <= MaxVal) {
    tree[idx] += val;
    idx += (idx & -idx);
  }
}
void SetA() {
  for (long long i = 0; i < n; i++) {
    a[i] = read(n - List[i].Num + 1);
    update(n - List[i].Num + 1, 1);
  }
}
void SetB() {
  for (long long i = n - 1; i >= 0; i--) {
    b[i] = read(List[i].Num);
    update(List[i].Num, 1);
  }
}
void Output() {
  long long t = 0;
  for (long long i = 0; i < n; i++) {
    t += a[i] * b[i];
  }
  cout << t << endl;
}
int main() {
  Input();
  SetA();
  for (long long i = 0; i < 1001000; i++) tree[i] = 0;
  SetB();
  Output();
  return 0;
}

```