#include <bits/stdc++.h>
using namespace std;
vector<long long> fb(1000010 + 1, 0), fe(1000010 + 1, 0);
void update(long long which, long long idx, long long val) {
  if (which == 1) {
    while (idx < 1000010) {
      fb[idx] += val;
      idx += idx & -idx;
    }
  } else {
    while (idx < 1000010) {
      fe[idx] += val;
      idx += idx & -idx;
    }
  }
}
long long query(long long which, long long idx) {
  if (which == 1) {
    long long ans = 0;
    while (idx > 0) {
      ans += fb[idx];
      idx -= idx & -idx;
    }
    return ans;
  } else {
    long long ans = 0;
    while (idx > 0) {
      ans += fe[idx];
      idx -= idx & -idx;
    }
    return ans;
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  long long begtime = clock();
  ;
  map<long long, long long> comp;
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (long long i = 0; i < n; ++i) cin >> arr[i];
  vector<long long> arr2 = arr;
  sort(arr2.begin(), arr2.end());
  long long cur = 1;
  for (long long i = 0; i < n; i += 1) comp[arr2[i]] = cur, cur += 1;
  long long ans = 0;
  for (long long i = 0; i < n; i += 1) {
    update(2, comp[arr[i]], 1);
  }
  for (long long i = 0; i < n; i += 1) {
    long long le = query(2, comp[arr[i]] - 1);
    long long gr = query(1, 1000010) - query(1, comp[arr[i]]);
    ans += le * gr;
    update(1, comp[arr[i]], 1);
    update(2, comp[arr[i]], -1);
  }
  cout << ans << endl;
  long long endtime = clock();
  cerr << endl
       << "Time elapsed: " << (endtime - begtime) * 1000 / CLOCKS_PER_SEC
       << " ms";
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
vector<long long> fb(1000010 + 1, 0), fe(1000010 + 1, 0);
void update(long long which, long long idx, long long val) {
  if (which == 1) {
    while (idx < 1000010) {
      fb[idx] += val;
      idx += idx & -idx;
    }
  } else {
    while (idx < 1000010) {
      fe[idx] += val;
      idx += idx & -idx;
    }
  }
}
long long query(long long which, long long idx) {
  if (which == 1) {
    long long ans = 0;
    while (idx > 0) {
      ans += fb[idx];
      idx -= idx & -idx;
    }
    return ans;
  } else {
    long long ans = 0;
    while (idx > 0) {
      ans += fe[idx];
      idx -= idx & -idx;
    }
    return ans;
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  long long begtime = clock();
  ;
  map<long long, long long> comp;
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (long long i = 0; i < n; ++i) cin >> arr[i];
  vector<long long> arr2 = arr;
  sort(arr2.begin(), arr2.end());
  long long cur = 1;
  for (long long i = 0; i < n; i += 1) comp[arr2[i]] = cur, cur += 1;
  long long ans = 0;
  for (long long i = 0; i < n; i += 1) {
    update(2, comp[arr[i]], 1);
  }
  for (long long i = 0; i < n; i += 1) {
    long long le = query(2, comp[arr[i]] - 1);
    long long gr = query(1, 1000010) - query(1, comp[arr[i]]);
    ans += le * gr;
    update(1, comp[arr[i]], 1);
    update(2, comp[arr[i]], -1);
  }
  cout << ans << endl;
  long long endtime = clock();
  cerr << endl
       << "Time elapsed: " << (endtime - begtime) * 1000 / CLOCKS_PER_SEC
       << " ms";
  return 0;
}

```