#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 50;
int a[N];
long long BIT[N], BIT2[N];
void update(int idx, long long val, long long FT[] = ::BIT) {
  while (idx < N) {
    FT[idx] += val;
    idx += (idx & -idx);
  }
}
long long query(int idx, long long FT[] = ::BIT) {
  long long ans = 0;
  while (idx > 0) {
    ans += FT[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
void compress(int n) {
  int k = 0;
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    m[a[i]];
  }
  for (auto &u : m) {
    u.second = ++k;
  }
  for (int i = 0; i < n; i++) {
    a[i] = m[a[i]];
  }
}
int main(int argc, const char **argv) {
  if (argc >= 2) {
    freopen(argv[1], "r", stdin);
  }
  int n;
  cin >> n;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  compress(n);
  for (int i = 0; i < n; i++) {
    ans += query(N, BIT2) - query(a[i], BIT2);
    update(a[i], query(N, BIT) - query(a[i], BIT), BIT2);
    update(a[i], 1, BIT);
  }
  cout << ans << endl;
}

### 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;
const int N = 1e6 + 50;
int a[N];
long long BIT[N], BIT2[N];
void update(int idx, long long val, long long FT[] = ::BIT) {
  while (idx < N) {
    FT[idx] += val;
    idx += (idx & -idx);
  }
}
long long query(int idx, long long FT[] = ::BIT) {
  long long ans = 0;
  while (idx > 0) {
    ans += FT[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
void compress(int n) {
  int k = 0;
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    m[a[i]];
  }
  for (auto &u : m) {
    u.second = ++k;
  }
  for (int i = 0; i < n; i++) {
    a[i] = m[a[i]];
  }
}
int main(int argc, const char **argv) {
  if (argc >= 2) {
    freopen(argv[1], "r", stdin);
  }
  int n;
  cin >> n;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  compress(n);
  for (int i = 0; i < n; i++) {
    ans += query(N, BIT2) - query(a[i], BIT2);
    update(a[i], query(N, BIT) - query(a[i], BIT), BIT2);
    update(a[i], 1, BIT);
  }
  cout << ans << endl;
}

```