#include <bits/stdc++.h>
using namespace std;
long long MOD = 1e9 + 7;
const long long int N = 1e6 + 10;
vector<long long int> a(N);
vector<long long int> BIT(N);
map<long long int, long long int> mp;
long long int n;
void init() {
  for (auto &x : BIT) x = 0;
}
void update(long long int x, long long int val) {
  for (; x < N; x += x & -x) {
    BIT[x] += val;
  }
}
long long int query(long long int x) {
  long long int sum = 0;
  for (; x > 0; x -= x & -x) sum += BIT[x];
  return sum;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  vector<long long int> b;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    b.push_back(a[i]);
  }
  sort(b.begin(), b.end());
  long long int val = 2;
  for (auto x : b) {
    if (!mp[x]) {
      mp[x] = val;
      val++;
    }
  }
  vector<long long int> v1(n), v2(n);
  for (long long int i = n - 1; i >= 0; i--) {
    long long int x = mp[a[i]];
    v1[i] = query(x - 1);
    update(x, 1);
  }
  init();
  for (long long int i = 0; i < n; i++) {
    long long int x = mp[a[i]];
    v2[i] = query(val) - query(x - 1);
    update(x, 1);
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) ans = ans + v1[i] * v2[i];
  cout << ans << endl;
}

### 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;
long long MOD = 1e9 + 7;
const long long int N = 1e6 + 10;
vector<long long int> a(N);
vector<long long int> BIT(N);
map<long long int, long long int> mp;
long long int n;
void init() {
  for (auto &x : BIT) x = 0;
}
void update(long long int x, long long int val) {
  for (; x < N; x += x & -x) {
    BIT[x] += val;
  }
}
long long int query(long long int x) {
  long long int sum = 0;
  for (; x > 0; x -= x & -x) sum += BIT[x];
  return sum;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  vector<long long int> b;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    b.push_back(a[i]);
  }
  sort(b.begin(), b.end());
  long long int val = 2;
  for (auto x : b) {
    if (!mp[x]) {
      mp[x] = val;
      val++;
    }
  }
  vector<long long int> v1(n), v2(n);
  for (long long int i = n - 1; i >= 0; i--) {
    long long int x = mp[a[i]];
    v1[i] = query(x - 1);
    update(x, 1);
  }
  init();
  for (long long int i = 0; i < n; i++) {
    long long int x = mp[a[i]];
    v2[i] = query(val) - query(x - 1);
    update(x, 1);
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) ans = ans + v1[i] * v2[i];
  cout << ans << endl;
}

```