#include <bits/stdc++.h>
using namespace std;
long long bit[4][1000006] = {0};
long long a[1000006], temp[1000006];
long long n, k;
void convert() {
  for (int i = 0; i < n; i++) {
    temp[i] = a[i];
  }
  sort(temp, temp + n);
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(temp, temp + n, a[i]) - temp + 1;
  }
}
void add(int t, int i, long long val) {
  for (; i <= n; i = i + (i & (-i))) {
    bit[t][i] = bit[t][i] + val;
  }
}
long long sum(int t, int i) {
  long long res = 0;
  for (; i > 0; i = i - (i & (-i))) {
    res = res + bit[t][i];
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  k = 3;
  long long i, l, x;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  convert();
  for (i = n - 1; i >= 0; i--) {
    x = a[i];
    add(1, x, 1LL);
    for (l = 1; l < k; l++) {
      add(l + 1, x, sum(l, x - 1));
    }
  }
  cout << sum(k, n);
  return 0;
}

### Prompt

Generate 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;
long long bit[4][1000006] = {0};
long long a[1000006], temp[1000006];
long long n, k;
void convert() {
  for (int i = 0; i < n; i++) {
    temp[i] = a[i];
  }
  sort(temp, temp + n);
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(temp, temp + n, a[i]) - temp + 1;
  }
}
void add(int t, int i, long long val) {
  for (; i <= n; i = i + (i & (-i))) {
    bit[t][i] = bit[t][i] + val;
  }
}
long long sum(int t, int i) {
  long long res = 0;
  for (; i > 0; i = i - (i & (-i))) {
    res = res + bit[t][i];
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  k = 3;
  long long i, l, x;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  convert();
  for (i = n - 1; i >= 0; i--) {
    x = a[i];
    add(1, x, 1LL);
    for (l = 1; l < k; l++) {
      add(l + 1, x, sum(l, x - 1));
    }
  }
  cout << sum(k, n);
  return 0;
}

```