#include <bits/stdc++.h>
using namespace std;
class TreeArray {
 private:
  int c[1000005];

 public:
  void init() { memset(c, 0, sizeof(c)); }
  int lowbit(int n) { return n & (-n); }
  void add(int pos, int n) {
    while (pos < 1000005) {
      c[pos] += n;
      pos += lowbit(pos);
    }
  }
  int getSum(int pos) {
    int ret = 0;
    while (pos) {
      ret += c[pos];
      pos -= lowbit(pos);
    }
    return ret;
  }
};
int num[1000005];
int tmp[1000005];
long long large[1000005];
long long small[1000005];
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &num[i]);
      tmp[i] = num[i];
    }
    sort(tmp, tmp + n);
    TreeArray obj;
    obj.init();
    for (int i = 0; i < n; i++) {
      int hash = lower_bound(tmp, tmp + n, num[i]) - tmp + 1;
      large[i] = i - obj.getSum(hash);
      obj.add(hash, 1);
    }
    obj.init();
    for (int i = n - 1; i >= 0; i--) {
      int hash = lower_bound(tmp, tmp + n, num[i]) - tmp + 1;
      small[i] = obj.getSum(hash);
      obj.add(hash, 1);
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      ans += large[i] * small[i];
    }
    cout << ans << endl;
  }
  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;
class TreeArray {
 private:
  int c[1000005];

 public:
  void init() { memset(c, 0, sizeof(c)); }
  int lowbit(int n) { return n & (-n); }
  void add(int pos, int n) {
    while (pos < 1000005) {
      c[pos] += n;
      pos += lowbit(pos);
    }
  }
  int getSum(int pos) {
    int ret = 0;
    while (pos) {
      ret += c[pos];
      pos -= lowbit(pos);
    }
    return ret;
  }
};
int num[1000005];
int tmp[1000005];
long long large[1000005];
long long small[1000005];
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &num[i]);
      tmp[i] = num[i];
    }
    sort(tmp, tmp + n);
    TreeArray obj;
    obj.init();
    for (int i = 0; i < n; i++) {
      int hash = lower_bound(tmp, tmp + n, num[i]) - tmp + 1;
      large[i] = i - obj.getSum(hash);
      obj.add(hash, 1);
    }
    obj.init();
    for (int i = n - 1; i >= 0; i--) {
      int hash = lower_bound(tmp, tmp + n, num[i]) - tmp + 1;
      small[i] = obj.getSum(hash);
      obj.add(hash, 1);
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      ans += large[i] * small[i];
    }
    cout << ans << endl;
  }
  return 0;
}

```