#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int INF = 1000000007;
const int MAXN = (int)1e6 + 1;
pair<int, int> power[MAXN];
long long L[MAXN], R[MAXN], L_BIT[MAXN], R_BIT[MAXN];
int n;
bool comp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
void update(long long arr[], int pos, long long val) {
  for (; pos <= n; pos += (pos & -pos)) {
    arr[pos] += val;
  }
}
long long query(long long arr[], int pos) {
  long long sum = 0;
  for (; pos > 0; pos -= (pos & -pos)) {
    sum += arr[pos];
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> power[i].first;
    power[i].second = i;
  }
  sort(power, power + n);
  for (int i = 0; i < n; i++) {
    power[i].first = i + 1;
  }
  sort(power, power + n, comp);
  for (int i = 0; i < n; i++) {
    L[i] = query(L_BIT, n) - query(L_BIT, power[i].first);
    update(L_BIT, power[i].first, 1);
  }
  for (int i = n - 1; i >= 0; i--) {
    R[i] = query(R_BIT, power[i].first - 1);
    update(R_BIT, power[i].first, 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += L[i] * R[i];
  }
  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 MOD = 1000000007;
const int INF = 1000000007;
const int MAXN = (int)1e6 + 1;
pair<int, int> power[MAXN];
long long L[MAXN], R[MAXN], L_BIT[MAXN], R_BIT[MAXN];
int n;
bool comp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
void update(long long arr[], int pos, long long val) {
  for (; pos <= n; pos += (pos & -pos)) {
    arr[pos] += val;
  }
}
long long query(long long arr[], int pos) {
  long long sum = 0;
  for (; pos > 0; pos -= (pos & -pos)) {
    sum += arr[pos];
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> power[i].first;
    power[i].second = i;
  }
  sort(power, power + n);
  for (int i = 0; i < n; i++) {
    power[i].first = i + 1;
  }
  sort(power, power + n, comp);
  for (int i = 0; i < n; i++) {
    L[i] = query(L_BIT, n) - query(L_BIT, power[i].first);
    update(L_BIT, power[i].first, 1);
  }
  for (int i = n - 1; i >= 0; i--) {
    R[i] = query(R_BIT, power[i].first - 1);
    update(R_BIT, power[i].first, 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += L[i] * R[i];
  }
  cout << ans;
  return 0;
}

```