#include <bits/stdc++.h>
using namespace std;
const int MaxN = (int)1e6 + 100;
long long tree[MaxN][2];
long long sum(int x, int y) {
  long long ans = 0;
  for (; x >= 0; x = (x & (x + 1)) - 1) ans += tree[x][y];
  return ans;
}
void update(int x, int val, int y) {
  for (; x <= MaxN; x |= (x + 1)) tree[x][y] += val;
}
int main() {
  if (1) {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  }
  if (0) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  int n;
  vector<int> v, s;
  map<int, int> mp;
  cin >> n;
  v.resize(n);
  s.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    s[i] = v[i];
  }
  sort(s.begin(), s.end());
  for (int i = 0; i < n; i++) mp[s[i]] = i;
  for (int i = 0; i < n; i++) v[i] = mp[v[i]];
  reverse(v.begin(), v.end());
  long long ans = 0;
  for (auto i : v)
    if (0) cerr << i << ' ';
  if (0) cerr << '\n';
  for (int i = 0; i < n; i++) {
    long long tm = sum(v[i], 0);
    update(v[i], 1, 0);
    update(v[i], tm, 1);
    ans += sum(v[i] - 1, 1);
    if (0) cerr << tm << ' ' << ans << '\n';
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = (int)1e6 + 100;
long long tree[MaxN][2];
long long sum(int x, int y) {
  long long ans = 0;
  for (; x >= 0; x = (x & (x + 1)) - 1) ans += tree[x][y];
  return ans;
}
void update(int x, int val, int y) {
  for (; x <= MaxN; x |= (x + 1)) tree[x][y] += val;
}
int main() {
  if (1) {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  }
  if (0) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  int n;
  vector<int> v, s;
  map<int, int> mp;
  cin >> n;
  v.resize(n);
  s.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    s[i] = v[i];
  }
  sort(s.begin(), s.end());
  for (int i = 0; i < n; i++) mp[s[i]] = i;
  for (int i = 0; i < n; i++) v[i] = mp[v[i]];
  reverse(v.begin(), v.end());
  long long ans = 0;
  for (auto i : v)
    if (0) cerr << i << ' ';
  if (0) cerr << '\n';
  for (int i = 0; i < n; i++) {
    long long tm = sum(v[i], 0);
    update(v[i], 1, 0);
    update(v[i], tm, 1);
    ans += sum(v[i] - 1, 1);
    if (0) cerr << tm << ' ' << ans << '\n';
  }
  cout << ans;
  return 0;
}

```