#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1000024;
int c[MAX_N], b[MAX_N], arr[MAX_N], lager[MAX_N];
void add(int x, int v) {
  for (; x < MAX_N; x += x & (-x)) c[x] += v;
}
long long getsum(int x) {
  long long ans = 0;
  for (; x; x -= x & (-x)) ans += c[x];
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
    b[i] = arr[i];
  }
  sort(b + 1, b + 1 + n);
  int num = unique(b + 1, b + 1 + n) - b - 1;
  for (int i = 1; i <= n; ++i) {
    arr[i] = lower_bound(b + 1, b + 1 + num, arr[i]) - b;
    add(arr[i], 1);
  }
  for (int i = 1; i <= n; ++i) {
    lager[i] = n - getsum(arr[i]);
  }
  memset(c, 0, sizeof(c));
  long long ans = 0;
  for (int i = n; i >= 1; --i) {
    long long tmp = getsum(arr[i]);
    long long tmp_ = lager[i] - (n - i - getsum(arr[i]));
    ;
    ;
    ;
    ;
    ;
    ;
    ans += tmp_ * tmp;
    add(arr[i], 1);
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int MAX_N = 1000024;
int c[MAX_N], b[MAX_N], arr[MAX_N], lager[MAX_N];
void add(int x, int v) {
  for (; x < MAX_N; x += x & (-x)) c[x] += v;
}
long long getsum(int x) {
  long long ans = 0;
  for (; x; x -= x & (-x)) ans += c[x];
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
    b[i] = arr[i];
  }
  sort(b + 1, b + 1 + n);
  int num = unique(b + 1, b + 1 + n) - b - 1;
  for (int i = 1; i <= n; ++i) {
    arr[i] = lower_bound(b + 1, b + 1 + num, arr[i]) - b;
    add(arr[i], 1);
  }
  for (int i = 1; i <= n; ++i) {
    lager[i] = n - getsum(arr[i]);
  }
  memset(c, 0, sizeof(c));
  long long ans = 0;
  for (int i = n; i >= 1; --i) {
    long long tmp = getsum(arr[i]);
    long long tmp_ = lager[i] - (n - i - getsum(arr[i]));
    ;
    ;
    ;
    ;
    ;
    ;
    ans += tmp_ * tmp;
    add(arr[i], 1);
  }
  printf("%I64d\n", ans);
  return 0;
}

```