#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-9;
const int mod = 1e9 + 7;
const long long INF = 1e18 + 1;
const long long inf = 1e12 + 1;
const int N = 2e6 + 9;
long long n;
long long ans;
long long a[N];
long long t[4 * N];
map<int, int> f;
vector<long long> g;
inline void update(int v, int l, int r, int pos) {
  if (l == r) {
    t[v]++;
    return;
  }
  int md = (l + r) / 2;
  if (pos <= md)
    update(v + v, l, md, pos);
  else
    update(v + v + 1, md + 1, r, pos);
  t[v] = t[v + v] + t[v + v + 1];
}
long long sum(int v, int l, int r, long long L, long long R) {
  if (l > R || r < L) return 0;
  if (L <= l && r <= R) return t[v];
  int md = (l + r) / 2;
  return sum(v + v, l, md, L, R) + sum(v + v + 1, md + 1, r, L, R);
}
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];
    g.push_back(a[i]);
  }
  sort(g.begin(), g.end());
  for (int i = 0; i < (long long)(g.size()); ++i) {
    f[g[i]] = i + 1;
  }
  update(1, 1, n, f[a[1]]);
  for (int i = 2; i < n; ++i) {
    long long q = sum(1, 1, n, f[a[i]], n);
    long long w = (f[a[i]] - 1) - sum(1, 1, n, 1, f[a[i]]);
    ans += (q * w);
    update(1, 1, n, f[a[i]]);
  }
  cout << ans;
  exit(0);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 double pi = acos(-1.0);
const double eps = 1e-9;
const int mod = 1e9 + 7;
const long long INF = 1e18 + 1;
const long long inf = 1e12 + 1;
const int N = 2e6 + 9;
long long n;
long long ans;
long long a[N];
long long t[4 * N];
map<int, int> f;
vector<long long> g;
inline void update(int v, int l, int r, int pos) {
  if (l == r) {
    t[v]++;
    return;
  }
  int md = (l + r) / 2;
  if (pos <= md)
    update(v + v, l, md, pos);
  else
    update(v + v + 1, md + 1, r, pos);
  t[v] = t[v + v] + t[v + v + 1];
}
long long sum(int v, int l, int r, long long L, long long R) {
  if (l > R || r < L) return 0;
  if (L <= l && r <= R) return t[v];
  int md = (l + r) / 2;
  return sum(v + v, l, md, L, R) + sum(v + v + 1, md + 1, r, L, R);
}
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];
    g.push_back(a[i]);
  }
  sort(g.begin(), g.end());
  for (int i = 0; i < (long long)(g.size()); ++i) {
    f[g[i]] = i + 1;
  }
  update(1, 1, n, f[a[1]]);
  for (int i = 2; i < n; ++i) {
    long long q = sum(1, 1, n, f[a[i]], n);
    long long w = (f[a[i]] - 1) - sum(1, 1, n, 1, f[a[i]]);
    ans += (q * w);
    update(1, 1, n, f[a[i]]);
  }
  cout << ans;
  exit(0);
}

```