#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
long long ft[maxn + 1];
int n;
int lso(int i) { return (i & (-i)); }
void update(int pos, int val) {
  if (pos == 0) return;
  for (pos = pos; pos <= n; pos += lso(pos)) ft[pos] += val;
}
long long query(int pos) {
  long long ans = ft[pos];
  for (pos = pos - lso(pos); pos > 0; pos -= lso(pos)) ans += ft[pos];
  return ans;
}
long long intervalo(int l, int r) { return query(r) - query(l - 1); }
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int q, ent;
  cin >> n;
  vector<pair<int, int> > a(n);
  vector<int> vet(n);
  for (int i = 0; i < n; i++) {
    cin >> ent;
    a[i] = {ent, i};
  }
  sort(a.begin(), a.end());
  for (int i = 0; i < n; i++) {
    vet[a[i].second] = i + 1;
  }
  reverse(vet.begin(), vet.end());
  vector<long long> inv1(n);
  vector<long long> inv2(n);
  int numAt;
  long long inv = 0;
  memset(ft, 0, sizeof ft);
  for (int i = 0; i < n; i++) {
    numAt = vet[i];
    inv = intervalo(1, numAt);
    inv1[i] = inv;
    update(numAt, 1);
  }
  inv = 0;
  memset(ft, 0, sizeof ft);
  for (int i = n - 1; i >= 0; i--) {
    numAt = vet[i];
    inv = intervalo(numAt, n);
    inv2[i] = inv;
    update(numAt, 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += inv1[i] * inv2[i];
  }
  cout << ans << endl;
}

### 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;
const int maxn = 1e6 + 10;
long long ft[maxn + 1];
int n;
int lso(int i) { return (i & (-i)); }
void update(int pos, int val) {
  if (pos == 0) return;
  for (pos = pos; pos <= n; pos += lso(pos)) ft[pos] += val;
}
long long query(int pos) {
  long long ans = ft[pos];
  for (pos = pos - lso(pos); pos > 0; pos -= lso(pos)) ans += ft[pos];
  return ans;
}
long long intervalo(int l, int r) { return query(r) - query(l - 1); }
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int q, ent;
  cin >> n;
  vector<pair<int, int> > a(n);
  vector<int> vet(n);
  for (int i = 0; i < n; i++) {
    cin >> ent;
    a[i] = {ent, i};
  }
  sort(a.begin(), a.end());
  for (int i = 0; i < n; i++) {
    vet[a[i].second] = i + 1;
  }
  reverse(vet.begin(), vet.end());
  vector<long long> inv1(n);
  vector<long long> inv2(n);
  int numAt;
  long long inv = 0;
  memset(ft, 0, sizeof ft);
  for (int i = 0; i < n; i++) {
    numAt = vet[i];
    inv = intervalo(1, numAt);
    inv1[i] = inv;
    update(numAt, 1);
  }
  inv = 0;
  memset(ft, 0, sizeof ft);
  for (int i = n - 1; i >= 0; i--) {
    numAt = vet[i];
    inv = intervalo(numAt, n);
    inv2[i] = inv;
    update(numAt, 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += inv1[i] * inv2[i];
  }
  cout << ans << endl;
}

```