#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 9;
vector<int> t(N * 4, 0), v;
int n, a[N];
int l[N], g[N];
void upd(int pos, int v = 1, int tl = 1, int tr = n) {
  if (tl == tr) {
    t[v] = 1;
    return;
  }
  int tm = tl + tr >> 1;
  if (pos <= tm)
    upd(pos, v + v, tl, tm);
  else
    upd(pos, v + v + 1, tm + 1, tr);
  t[v] = t[v + v] + t[v + v + 1];
}
int get(int l, int r, int v = 1, int tl = 1, int tr = n) {
  if (l <= tl && tr <= r) return t[v];
  if (r < tl || tr < l) return 0;
  int tm = tl + tr >> 1;
  return get(l, r, v + v, tl, tm) + get(l, r, v + v + 1, tm + 1, tr);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  unique(v.begin(), v.end());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  for (int i = 1; i <= n; i++) {
    g[i] = get(a[i], n);
    upd(a[i]);
  }
  t.assign(N * 4, 0);
  for (int i = n; i >= 1; i--) {
    l[i] = get(1, a[i]);
    upd(a[i]);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += (1LL * g[i]) * (1LL * l[i]);
  }
  cout << ans;
}

### Prompt

In cpp, your task is to solve 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 + 9;
vector<int> t(N * 4, 0), v;
int n, a[N];
int l[N], g[N];
void upd(int pos, int v = 1, int tl = 1, int tr = n) {
  if (tl == tr) {
    t[v] = 1;
    return;
  }
  int tm = tl + tr >> 1;
  if (pos <= tm)
    upd(pos, v + v, tl, tm);
  else
    upd(pos, v + v + 1, tm + 1, tr);
  t[v] = t[v + v] + t[v + v + 1];
}
int get(int l, int r, int v = 1, int tl = 1, int tr = n) {
  if (l <= tl && tr <= r) return t[v];
  if (r < tl || tr < l) return 0;
  int tm = tl + tr >> 1;
  return get(l, r, v + v, tl, tm) + get(l, r, v + v + 1, tm + 1, tr);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  unique(v.begin(), v.end());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  for (int i = 1; i <= n; i++) {
    g[i] = get(a[i], n);
    upd(a[i]);
  }
  t.assign(N * 4, 0);
  for (int i = n; i >= 1; i--) {
    l[i] = get(1, a[i]);
    upd(a[i]);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += (1LL * g[i]) * (1LL * l[i]);
  }
  cout << ans;
}

```