#include <bits/stdc++.h>
using namespace std;
const int mN = 1e6 + 100;
long long BIT[2][mN];
long long get(int t, int i) {
  ++i;
  long long ret = 0;
  while (i) {
    ret += BIT[t][i];
    i -= i & -i;
  }
  return ret;
}
void add(int t, int i, long long v) {
  ++i;
  while (i < mN) {
    BIT[t][i] += v;
    i += i & -i;
  }
}
int n;
long long arr[mN];
vector<int> vec;
int com(int v) { return lower_bound(vec.begin(), vec.end(), v) - vec.begin(); }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lld", &arr[i]), vec.push_back(arr[i]);
  sort(vec.begin(), vec.end());
  long long out = 0;
  for (int i = 0; i < n; ++i) {
    out += get(1, n - 1 - com(arr[i]) - 1);
    add(0, n - 1 - com(arr[i]), 1);
    add(1, n - 1 - com(arr[i]), get(0, n - 1 - com(arr[i]) - 1));
  }
  printf("%lld\n", out);
}

### 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;
const int mN = 1e6 + 100;
long long BIT[2][mN];
long long get(int t, int i) {
  ++i;
  long long ret = 0;
  while (i) {
    ret += BIT[t][i];
    i -= i & -i;
  }
  return ret;
}
void add(int t, int i, long long v) {
  ++i;
  while (i < mN) {
    BIT[t][i] += v;
    i += i & -i;
  }
}
int n;
long long arr[mN];
vector<int> vec;
int com(int v) { return lower_bound(vec.begin(), vec.end(), v) - vec.begin(); }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lld", &arr[i]), vec.push_back(arr[i]);
  sort(vec.begin(), vec.end());
  long long out = 0;
  for (int i = 0; i < n; ++i) {
    out += get(1, n - 1 - com(arr[i]) - 1);
    add(0, n - 1 - com(arr[i]), 1);
    add(1, n - 1 - com(arr[i]), get(0, n - 1 - com(arr[i]) - 1));
  }
  printf("%lld\n", out);
}

```