#include <bits/stdc++.h>
using namespace std;
vector<long long int> bit1(1000001, 0);
vector<long long int> bit2(1000001, 0);
void update1(long long int x, long long int val) {
  for (; x <= 1000000; x += (x & (-x))) {
    bit1[x] += val;
  }
}
long long int query1(long long int x) {
  long long int sum = 0;
  for (; x > 0; x -= (x & (-x))) {
    sum += bit1[x];
  }
  return sum;
}
void update2(long long int x, long long int val) {
  for (; x <= 1000000; x += (x & (-x))) {
    bit2[x] += val;
  }
}
long long int query2(long long int x) {
  long long int sum = 0;
  for (; x > 0; x -= (x & (-x))) {
    sum += bit2[x];
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  vector<long long int> u = v;
  sort((u).begin(), (u).end());
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) {
    if (m.find(u[i]) == m.end()) {
      m[u[i]] = (i + 1);
    }
  }
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) {
    update1(m[v[i]], 1);
    long long int p = query1(m[v[i]]);
    a[i] = (i + 1) - p;
  }
  vector<long long int> b(n);
  for (long long int i = (n - 1); i >= 0; i--) {
    update2(m[v[i]], 1);
    long long int p = query2(m[v[i]] - 1);
    b[i] = p;
  }
  long long int c = 0;
  for (long long int i = 1; i < (n - 1); i++) {
    c += (a[i] * b[i]);
  }
  cout << c << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 int> bit1(1000001, 0);
vector<long long int> bit2(1000001, 0);
void update1(long long int x, long long int val) {
  for (; x <= 1000000; x += (x & (-x))) {
    bit1[x] += val;
  }
}
long long int query1(long long int x) {
  long long int sum = 0;
  for (; x > 0; x -= (x & (-x))) {
    sum += bit1[x];
  }
  return sum;
}
void update2(long long int x, long long int val) {
  for (; x <= 1000000; x += (x & (-x))) {
    bit2[x] += val;
  }
}
long long int query2(long long int x) {
  long long int sum = 0;
  for (; x > 0; x -= (x & (-x))) {
    sum += bit2[x];
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  vector<long long int> u = v;
  sort((u).begin(), (u).end());
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) {
    if (m.find(u[i]) == m.end()) {
      m[u[i]] = (i + 1);
    }
  }
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) {
    update1(m[v[i]], 1);
    long long int p = query1(m[v[i]]);
    a[i] = (i + 1) - p;
  }
  vector<long long int> b(n);
  for (long long int i = (n - 1); i >= 0; i--) {
    update2(m[v[i]], 1);
    long long int p = query2(m[v[i]] - 1);
    b[i] = p;
  }
  long long int c = 0;
  for (long long int i = 1; i < (n - 1); i++) {
    c += (a[i] * b[i]);
  }
  cout << c << endl;
}

```