#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
template <class T>
typename T::value_type arr_sum(const T& v, int n) {
  typename T::value_type sum = 0;
  for (int i = (0); i < (n); ++i) sum += v[i];
  return sum;
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
const int MAXN = 2000000;
vector<int> t[4 * MAXN];
void build(const vector<int>& a, int v, int tl, int tr) {
  if (tl == tr)
    t[v] = vector<int>(1, a[tl]);
  else {
    int tm = (tl + tr) / 2;
    build(a, v * 2 + 1, tl, tm);
    build(a, v * 2 + 2, tm + 1, tr);
    merge(t[v * 2 + 1].begin(), t[v * 2 + 1].end(), t[v * 2 + 2].begin(),
          t[v * 2 + 2].end(), back_inserter(t[v]));
  }
}
int query(int v, int tl, int tr, int l, int r, int x) {
  if (l > r) return 0;
  if (l == tl && tr == r) {
    auto pos = lower_bound(t[v].begin(), t[v].end(), x);
    return pos - t[v].begin();
  }
  int tm = (tl + tr) / 2;
  return query(v * 2 + 1, tl, tm, l, min(r, tm), x) +
         query(v * 2 + 2, tm + 1, tr, max(l, tm + 1), r, x);
}
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int __i = 0; __i < (n); ++__i) cin >> v[__i];
  ;
  build(v, 0, 0, n - 1);
  long long res = 0;
  for (int i = (0); i < (n); ++i) {
    long long tres = i - query(0, 0, n - 1, 0, i - 1, v[i] + 1);
    tres *= query(0, 0, n - 1, i + 1, n - 1, v[i]);
    res += tres;
  }
  cout << res;
}

### Prompt

Please formulate 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;
template <class T>
typename T::value_type arr_sum(const T& v, int n) {
  typename T::value_type sum = 0;
  for (int i = (0); i < (n); ++i) sum += v[i];
  return sum;
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
const int MAXN = 2000000;
vector<int> t[4 * MAXN];
void build(const vector<int>& a, int v, int tl, int tr) {
  if (tl == tr)
    t[v] = vector<int>(1, a[tl]);
  else {
    int tm = (tl + tr) / 2;
    build(a, v * 2 + 1, tl, tm);
    build(a, v * 2 + 2, tm + 1, tr);
    merge(t[v * 2 + 1].begin(), t[v * 2 + 1].end(), t[v * 2 + 2].begin(),
          t[v * 2 + 2].end(), back_inserter(t[v]));
  }
}
int query(int v, int tl, int tr, int l, int r, int x) {
  if (l > r) return 0;
  if (l == tl && tr == r) {
    auto pos = lower_bound(t[v].begin(), t[v].end(), x);
    return pos - t[v].begin();
  }
  int tm = (tl + tr) / 2;
  return query(v * 2 + 1, tl, tm, l, min(r, tm), x) +
         query(v * 2 + 2, tm + 1, tr, max(l, tm + 1), r, x);
}
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int __i = 0; __i < (n); ++__i) cin >> v[__i];
  ;
  build(v, 0, 0, n - 1);
  long long res = 0;
  for (int i = (0); i < (n); ++i) {
    long long tres = i - query(0, 0, n - 1, 0, i - 1, v[i] + 1);
    tres *= query(0, 0, n - 1, i + 1, n - 1, v[i]);
    res += tres;
  }
  cout << res;
}

```