#include <bits/stdc++.h>
using namespace std;
long long BIT1[2000006], BIT2[2000006];
void update1(long long i, long long val) {
  for (; i < 2000006; i += (i & -i)) BIT1[i] += val;
}
long long query1(long long i) {
  long long sum = 0;
  for (; i > 0; i -= (i & -i)) sum += BIT1[i];
  return sum;
}
void update2(long long i, long long val) {
  for (; i < 2000006; i += (i & -i)) BIT2[i] += val;
}
long long query2(long long i) {
  long long sum = 0;
  for (; i > 0; i -= (i & -i)) sum += BIT2[i];
  return sum;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  vector<long long> tot(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    tot[i] = arr[i];
  }
  sort(tot.begin(), tot.end());
  for (int i = 0; i < n; i++) {
    arr[i] = lower_bound(tot.begin(), tot.end(), arr[i]) - tot.begin();
    arr[i]++;
  }
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (arr[i] > 1) ans += query2(arr[i] - 1);
    long long val = 0;
    if (arr[i] > 1) val = query1(arr[i] - 1);
    update2(arr[i], val);
    update1(arr[i], 1);
  }
  cout << ans;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
long long BIT1[2000006], BIT2[2000006];
void update1(long long i, long long val) {
  for (; i < 2000006; i += (i & -i)) BIT1[i] += val;
}
long long query1(long long i) {
  long long sum = 0;
  for (; i > 0; i -= (i & -i)) sum += BIT1[i];
  return sum;
}
void update2(long long i, long long val) {
  for (; i < 2000006; i += (i & -i)) BIT2[i] += val;
}
long long query2(long long i) {
  long long sum = 0;
  for (; i > 0; i -= (i & -i)) sum += BIT2[i];
  return sum;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  vector<long long> tot(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    tot[i] = arr[i];
  }
  sort(tot.begin(), tot.end());
  for (int i = 0; i < n; i++) {
    arr[i] = lower_bound(tot.begin(), tot.end(), arr[i]) - tot.begin();
    arr[i]++;
  }
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (arr[i] > 1) ans += query2(arr[i] - 1);
    long long val = 0;
    if (arr[i] > 1) val = query1(arr[i] - 1);
    update2(arr[i], val);
    update1(arr[i], 1);
  }
  cout << ans;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```