#include <bits/stdc++.h>
using namespace std;
const long long N = 5e6 + 5;
long long n, a[N], ans, f1[N], f2[N], s[N];
void Set(long long i, long long id = 1, long long l = 0, long long r = n) {
  if (r - l == 1) {
    s[id]++;
    return;
  }
  if (i < l or i >= r) return;
  if (i < ((l + r) / 2))
    Set(i, (id * 2), l, ((l + r) / 2));
  else
    Set(i, ((id * 2) + 1), ((l + r) / 2), r);
  s[id] = s[(id * 2)] + s[((id * 2) + 1)];
}
long long Get(long long L, long long R, long long id = 1, long long l = 0,
              long long r = n) {
  if (R <= l or r <= L) return 0;
  if (L <= l and r <= R) return s[id];
  return (Get(L, R, (id * 2), l, ((l + r) / 2)) +
          Get(L, R, ((id * 2) + 1), ((l + r) / 2), r));
}
void compress() {
  vector<long long> u;
  for (long long i = 0; i < n; i++) u.push_back(a[i]);
  sort(u.begin(), u.end());
  u.resize(unique(u.begin(), u.end()) - u.begin());
  for (long long i = 0; i < n; i++)
    a[i] = lower_bound(u.begin(), u.end(), a[i]) - u.begin();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  compress();
  for (long long i = n - 1; i >= 0; i--) {
    Set(a[i]);
    f1[i] = Get(0, a[i]);
  }
  fill(s, s + N, 0);
  for (long long i = 0; i < n; i++) {
    Set(a[i]);
    f2[i] = Get(a[i] + 1, n);
  }
  for (long long i = 0; i < n; i++) {
    ans += f1[i] * f2[i];
  }
  cout << ans;
}

### Prompt

Develop a solution in CPP 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;
const long long N = 5e6 + 5;
long long n, a[N], ans, f1[N], f2[N], s[N];
void Set(long long i, long long id = 1, long long l = 0, long long r = n) {
  if (r - l == 1) {
    s[id]++;
    return;
  }
  if (i < l or i >= r) return;
  if (i < ((l + r) / 2))
    Set(i, (id * 2), l, ((l + r) / 2));
  else
    Set(i, ((id * 2) + 1), ((l + r) / 2), r);
  s[id] = s[(id * 2)] + s[((id * 2) + 1)];
}
long long Get(long long L, long long R, long long id = 1, long long l = 0,
              long long r = n) {
  if (R <= l or r <= L) return 0;
  if (L <= l and r <= R) return s[id];
  return (Get(L, R, (id * 2), l, ((l + r) / 2)) +
          Get(L, R, ((id * 2) + 1), ((l + r) / 2), r));
}
void compress() {
  vector<long long> u;
  for (long long i = 0; i < n; i++) u.push_back(a[i]);
  sort(u.begin(), u.end());
  u.resize(unique(u.begin(), u.end()) - u.begin());
  for (long long i = 0; i < n; i++)
    a[i] = lower_bound(u.begin(), u.end(), a[i]) - u.begin();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  compress();
  for (long long i = n - 1; i >= 0; i--) {
    Set(a[i]);
    f1[i] = Get(0, a[i]);
  }
  fill(s, s + N, 0);
  for (long long i = 0; i < n; i++) {
    Set(a[i]);
    f2[i] = Get(a[i] + 1, n);
  }
  for (long long i = 0; i < n; i++) {
    ans += f1[i] * f2[i];
  }
  cout << ans;
}

```