#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e6;
long long ft[5][MAXN + 5];
long long memo[5][MAXN + 5], getId[MAXN + 5];
long long n, arr[MAXN + 5];
pair<long long, long long> sorted[MAXN + 5];
void update(long long id, long long idx, long long val) {
  for (long long i = idx; i <= n; i += (i & (-i))) ft[id][i] += val;
  return;
}
long long get(long long id, long long idx) {
  long long ret = 0;
  for (long long i = idx; i > 0; i -= (i & (-i))) ret += ft[id][i];
  return ret;
}
void solve() {
  long long ans = 0;
  for (long long i = 1; i <= n; i++) memo[1][i] = 1;
  for (long long i = 2; i <= 3; i++) {
    for (long long j = 1; j <= n; j++) {
      update(i - 1, getId[j], memo[i - 1][j]);
      memo[i][j] = get(i - 1, getId[j] - 1);
      if (i == 3) ans += memo[i][j];
    }
  }
  cout << ans << endl;
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first > b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(memo, -1, sizeof(memo));
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> arr[i];
    sorted[i] = make_pair(arr[i], i);
  }
  sort(sorted + 1, sorted + n + 1, cmp);
  for (long long i = 1; i <= n; i++) getId[i] = sorted[i].second;
  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;
const long long MAXN = 1e6;
long long ft[5][MAXN + 5];
long long memo[5][MAXN + 5], getId[MAXN + 5];
long long n, arr[MAXN + 5];
pair<long long, long long> sorted[MAXN + 5];
void update(long long id, long long idx, long long val) {
  for (long long i = idx; i <= n; i += (i & (-i))) ft[id][i] += val;
  return;
}
long long get(long long id, long long idx) {
  long long ret = 0;
  for (long long i = idx; i > 0; i -= (i & (-i))) ret += ft[id][i];
  return ret;
}
void solve() {
  long long ans = 0;
  for (long long i = 1; i <= n; i++) memo[1][i] = 1;
  for (long long i = 2; i <= 3; i++) {
    for (long long j = 1; j <= n; j++) {
      update(i - 1, getId[j], memo[i - 1][j]);
      memo[i][j] = get(i - 1, getId[j] - 1);
      if (i == 3) ans += memo[i][j];
    }
  }
  cout << ans << endl;
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first > b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(memo, -1, sizeof(memo));
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> arr[i];
    sorted[i] = make_pair(arr[i], i);
  }
  sort(sorted + 1, sorted + n + 1, cmp);
  for (long long i = 1; i <= n; i++) getId[i] = sorted[i].second;
  solve();
  return 0;
}

```