#include <bits/stdc++.h>
using namespace std;
long long n;
long long arr[1000003];
void read() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> arr[i];
}
vector<long long> v;
map<long long, long long> mp;
long long first[1000003], second[1000003];
void update1(long long pos, long long val) {
  for (pos; pos <= n; pos += pos & -pos) first[pos] += val;
}
void update2(long long pos, long long val) {
  for (pos; pos <= n; pos += pos & -pos) second[pos] += val;
}
long long get1(long long pos) {
  if (pos < 1) return 0;
  long long ans = 0;
  for (pos; pos >= 1; pos -= pos & -pos) ans += first[pos];
  return ans;
}
long long get2(long long pos) {
  if (pos < 1) return 0;
  long long ans = 0;
  for (pos; pos >= 1; pos -= pos & -pos) ans += second[pos];
  return ans;
}
void solve() {
  for (long long i = 1; i <= n; i++) v.push_back(arr[i]);
  sort(v.begin(), v.end());
  for (long long i = 1; i <= n; i++) mp[v[i - 1]] = i;
  for (long long i = 1; i <= n; i++) arr[i] = mp[arr[i]];
  long long ans = 0;
  for (long long i = 1; i <= n; i++) update2(arr[i], 1);
  for (long long i = 1; i <= n; i++) {
    update2(arr[i], -1);
    ans += (get1(n) - get1(arr[i])) * get2(arr[i] - 1);
    update1(arr[i], 1);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  read();
  solve();
}

### Prompt

In CPP, your task is to solve 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 n;
long long arr[1000003];
void read() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> arr[i];
}
vector<long long> v;
map<long long, long long> mp;
long long first[1000003], second[1000003];
void update1(long long pos, long long val) {
  for (pos; pos <= n; pos += pos & -pos) first[pos] += val;
}
void update2(long long pos, long long val) {
  for (pos; pos <= n; pos += pos & -pos) second[pos] += val;
}
long long get1(long long pos) {
  if (pos < 1) return 0;
  long long ans = 0;
  for (pos; pos >= 1; pos -= pos & -pos) ans += first[pos];
  return ans;
}
long long get2(long long pos) {
  if (pos < 1) return 0;
  long long ans = 0;
  for (pos; pos >= 1; pos -= pos & -pos) ans += second[pos];
  return ans;
}
void solve() {
  for (long long i = 1; i <= n; i++) v.push_back(arr[i]);
  sort(v.begin(), v.end());
  for (long long i = 1; i <= n; i++) mp[v[i - 1]] = i;
  for (long long i = 1; i <= n; i++) arr[i] = mp[arr[i]];
  long long ans = 0;
  for (long long i = 1; i <= n; i++) update2(arr[i], 1);
  for (long long i = 1; i <= n; i++) {
    update2(arr[i], -1);
    ans += (get1(n) - get1(arr[i])) * get2(arr[i] - 1);
    update1(arr[i], 1);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  read();
  solve();
}

```