#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100;
void remap(int* a, int n) {
  int p[N];
  for (int i = 0; i < n; ++i) p[i] = i;
  sort(p, p + n, [=](int x, int y) -> bool { return a[x] < a[y]; });
  for (int i = 0; i < n; ++i) a[p[i]] = i + 1;
}
int fw[N];
int query(int v) {
  int ret = 0;
  while (v > 0) {
    ret += fw[v];
    v -= (v & -v);
  }
  return ret;
}
void add(int v) {
  while (v < N) {
    ++fw[v];
    v += (v & -v);
  }
}
int a[N];
int le[N], rg[N];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  remap(a, n);
  for (int i = 0; i < n; ++i) {
    int smallerle = query(a[i]);
    le[i] = i - smallerle;
    rg[i] = a[i] - 1 - smallerle;
    add(a[i]);
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += le[i] * 1LL * rg[i];
  }
  cout << ans << '\n';
}

### 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 N = 1e6 + 100;
void remap(int* a, int n) {
  int p[N];
  for (int i = 0; i < n; ++i) p[i] = i;
  sort(p, p + n, [=](int x, int y) -> bool { return a[x] < a[y]; });
  for (int i = 0; i < n; ++i) a[p[i]] = i + 1;
}
int fw[N];
int query(int v) {
  int ret = 0;
  while (v > 0) {
    ret += fw[v];
    v -= (v & -v);
  }
  return ret;
}
void add(int v) {
  while (v < N) {
    ++fw[v];
    v += (v & -v);
  }
}
int a[N];
int le[N], rg[N];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  remap(a, n);
  for (int i = 0; i < n; ++i) {
    int smallerle = query(a[i]);
    le[i] = i - smallerle;
    rg[i] = a[i] - 1 - smallerle;
    add(a[i]);
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += le[i] * 1LL * rg[i];
  }
  cout << ans << '\n';
}

```