#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
int a[N];
vector<int> v;
long long cont[N << 2];
long long g[N], li[N];
void push_up(int rt) { cont[rt] = cont[rt << 1] + cont[rt << 1 | 1]; }
void update(int l, int r, int rt, int id) {
  if (l == r) {
    cont[rt] = 1;
    return;
  }
  int m = (l + r) >> 1;
  if (m >= id)
    update(l, m, rt << 1, id);
  else
    update(m + 1, r, rt << 1 | 1, id);
  push_up(rt);
}
long long query(int l, int r, int rt, int L, int R) {
  if (L <= l && R >= r) return cont[rt];
  int m = (l + r) >> 1;
  long long res = 0;
  if (m >= L) res += query(l, m, rt << 1, L, R);
  if (m < R) res += query(m + 1, r, rt << 1 | 1, L, R);
  return res;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  memset(cont, 0, sizeof(cont));
  memset(g, 0, sizeof(g));
  int r = v.size() + 1;
  for (int i = 0; i < n; i++) {
    int id = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
    g[i] = query(1, r, 1, id, r);
    update(1, r, 1, id);
  }
  memset(cont, 0, sizeof(cont));
  for (int i = n - 1; i >= 0; i--) {
    int id = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
    li[i] = query(1, r, 1, 1, id);
    update(1, r, 1, id);
  }
  long long res = 0;
  for (int i = 0; i < n; i++) {
    res += g[i] * li[i];
  }
  cout << res << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 = 1000005;
int a[N];
vector<int> v;
long long cont[N << 2];
long long g[N], li[N];
void push_up(int rt) { cont[rt] = cont[rt << 1] + cont[rt << 1 | 1]; }
void update(int l, int r, int rt, int id) {
  if (l == r) {
    cont[rt] = 1;
    return;
  }
  int m = (l + r) >> 1;
  if (m >= id)
    update(l, m, rt << 1, id);
  else
    update(m + 1, r, rt << 1 | 1, id);
  push_up(rt);
}
long long query(int l, int r, int rt, int L, int R) {
  if (L <= l && R >= r) return cont[rt];
  int m = (l + r) >> 1;
  long long res = 0;
  if (m >= L) res += query(l, m, rt << 1, L, R);
  if (m < R) res += query(m + 1, r, rt << 1 | 1, L, R);
  return res;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  memset(cont, 0, sizeof(cont));
  memset(g, 0, sizeof(g));
  int r = v.size() + 1;
  for (int i = 0; i < n; i++) {
    int id = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
    g[i] = query(1, r, 1, id, r);
    update(1, r, 1, id);
  }
  memset(cont, 0, sizeof(cont));
  for (int i = n - 1; i >= 0; i--) {
    int id = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
    li[i] = query(1, r, 1, 1, id);
    update(1, r, 1, id);
  }
  long long res = 0;
  for (int i = 0; i < n; i++) {
    res += g[i] * li[i];
  }
  cout << res << endl;
  return 0;
}

```