#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
map<int, int> mp;
long long ans, fen[N];
int n, tmp, a[N], b[N], c[N];
long long get(int p) {
  long long res = 0;
  for (p++; p; p -= p & -p) res += fen[p];
  return res;
}
void add(int p, int x) {
  for (p++; p < N; p += p & -p) fen[p] += x;
}
void input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  mp[b[0]] = ++tmp;
  for (int i = 1; i < n; i++)
    if (b[i] != b[i - 1])
      if (!mp[b[i]]) mp[b[i]] = ++tmp;
}
void solve() {
  for (int i = n - 1; i >= 0; i--) {
    c[i] = get(mp[a[i]] - 1);
    add(mp[a[i]], 1);
  }
  memset(fen, 0, sizeof fen);
  for (int i = n - 1; i >= 0; i--) {
    ans += get(mp[a[i]] - 1);
    add(mp[a[i]], c[i]);
  }
}
void output() { cout << ans; }
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input(), solve(), output();
  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 int N = 1e6 + 10;
map<int, int> mp;
long long ans, fen[N];
int n, tmp, a[N], b[N], c[N];
long long get(int p) {
  long long res = 0;
  for (p++; p; p -= p & -p) res += fen[p];
  return res;
}
void add(int p, int x) {
  for (p++; p < N; p += p & -p) fen[p] += x;
}
void input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  mp[b[0]] = ++tmp;
  for (int i = 1; i < n; i++)
    if (b[i] != b[i - 1])
      if (!mp[b[i]]) mp[b[i]] = ++tmp;
}
void solve() {
  for (int i = n - 1; i >= 0; i--) {
    c[i] = get(mp[a[i]] - 1);
    add(mp[a[i]], 1);
  }
  memset(fen, 0, sizeof fen);
  for (int i = n - 1; i >= 0; i--) {
    ans += get(mp[a[i]] - 1);
    add(mp[a[i]], c[i]);
  }
}
void output() { cout << ans; }
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input(), solve(), output();
  return 0;
}

```