#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5;
int n, a[MAXN], p[MAXN];
int seg[MAXN * 4];
vector<pair<int, int> > vec;
void add(int ind, int val = 1, int id = 1, int b = 0, int e = n) {
  if (ind < b || ind >= e) return;
  if (b + 1 == e) {
    seg[id]++;
    return;
  }
  int mid = (b + e) / 2;
  if (ind < mid)
    add(ind, val, id * 2, b, mid);
  else
    add(ind, val, id * 2 + 1, mid, e);
  seg[id] = seg[id * 2] + seg[id * 2 + 1];
}
int get(int ind, int id = 1, int b = 0, int e = n) {
  int ans = 0;
  if (b >= ind) return 0;
  if (e <= ind) return seg[id];
  int mid = (b + e) / 2;
  ans += get(ind, id * 2, b, mid);
  ans += get(ind, id * 2 + 1, mid, e);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  set<pair<int, int> > s;
  for (int i = 0; i < n; i++) cin >> a[i], s.insert({a[i], i});
  int i = 0;
  for (set<pair<int, int> >::iterator it = s.begin(); it != s.end(); it++)
    p[it->second] = (long long)i, i++;
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    long long x = get(p[i]);
    add(p[i]);
    ans += x * (x + (long long)i - p[i]);
  }
  cout << ans;
  return 0;
}

### Prompt

Generate 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 MAXN = 1e6 + 5;
int n, a[MAXN], p[MAXN];
int seg[MAXN * 4];
vector<pair<int, int> > vec;
void add(int ind, int val = 1, int id = 1, int b = 0, int e = n) {
  if (ind < b || ind >= e) return;
  if (b + 1 == e) {
    seg[id]++;
    return;
  }
  int mid = (b + e) / 2;
  if (ind < mid)
    add(ind, val, id * 2, b, mid);
  else
    add(ind, val, id * 2 + 1, mid, e);
  seg[id] = seg[id * 2] + seg[id * 2 + 1];
}
int get(int ind, int id = 1, int b = 0, int e = n) {
  int ans = 0;
  if (b >= ind) return 0;
  if (e <= ind) return seg[id];
  int mid = (b + e) / 2;
  ans += get(ind, id * 2, b, mid);
  ans += get(ind, id * 2 + 1, mid, e);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  set<pair<int, int> > s;
  for (int i = 0; i < n; i++) cin >> a[i], s.insert({a[i], i});
  int i = 0;
  for (set<pair<int, int> >::iterator it = s.begin(); it != s.end(); it++)
    p[it->second] = (long long)i, i++;
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    long long x = get(p[i]);
    add(p[i]);
    ans += x * (x + (long long)i - p[i]);
  }
  cout << ans;
  return 0;
}

```