#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100, K = 5;
unsigned long long n, k, ans, a[N], dp[N][K], fen[N][K];
vector<int> vec;
unordered_map<int, int> REAL;
void add(int dex, int cnt, unsigned long long val) {
  for (; dex < N; dex += dex & -dex) fen[dex][cnt] += val;
}
unsigned long long sum(int dex, int cnt) {
  return (dex ? fen[dex][cnt] + sum(dex - (dex & -dex), cnt) : 0);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  k = 3;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vec.push_back(a[i]);
  }
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (int i = 0; i < vec.size(); i++) REAL[vec[i]] = i + 1;
  for (int i = 0; i < n; i++) a[i] = REAL[a[i]];
  reverse(a, a + n);
  for (int dex = 0; dex < n; dex++)
    for (int cnt = 1; cnt <= k; cnt++) {
      dp[dex][cnt] = (cnt == 1 ? 1 : sum(a[dex] - 1, cnt - 1));
      add(a[dex], cnt, dp[dex][cnt]);
    }
  for (int i = 0; i < n; i++) ans += dp[i][k];
  cout << ans;
  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;
const int N = 1e6 + 100, K = 5;
unsigned long long n, k, ans, a[N], dp[N][K], fen[N][K];
vector<int> vec;
unordered_map<int, int> REAL;
void add(int dex, int cnt, unsigned long long val) {
  for (; dex < N; dex += dex & -dex) fen[dex][cnt] += val;
}
unsigned long long sum(int dex, int cnt) {
  return (dex ? fen[dex][cnt] + sum(dex - (dex & -dex), cnt) : 0);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  k = 3;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vec.push_back(a[i]);
  }
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (int i = 0; i < vec.size(); i++) REAL[vec[i]] = i + 1;
  for (int i = 0; i < n; i++) a[i] = REAL[a[i]];
  reverse(a, a + n);
  for (int dex = 0; dex < n; dex++)
    for (int cnt = 1; cnt <= k; cnt++) {
      dp[dex][cnt] = (cnt == 1 ? 1 : sum(a[dex] - 1, cnt - 1));
      add(a[dex], cnt, dp[dex][cnt]);
    }
  for (int i = 0; i < n; i++) ans += dp[i][k];
  cout << ans;
  return 0;
}

```