#include <bits/stdc++.h>
using namespace std;
int static fast = []() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  return 0;
}();
class BinaryIndexTree {
 public:
  BinaryIndexTree(vector<int> nums) {
    int n = nums.size();
    arr = vector<int>(n, 0);
    bit_tree = vector<int>(n + 1, 0);
    for (int i = 0; i < n; i++) update(i, nums[i]);
  }
  void update(int idx, int val) {
    int delta = val - arr[idx];
    arr[idx] = val;
    int x = idx + 1;
    while (x < bit_tree.size()) {
      bit_tree[x] += delta;
      x += x & (-x);
    }
  }
  int _sumRange(int x) {
    int ret = 0;
    while (x > 0) {
      ret += bit_tree[x];
      x -= x & (-x);
    }
    return ret;
  }
  int sumRange(int from, int to) { return _sumRange(to + 1) - _sumRange(from); }
  int find_kth_smaller(int kth) {
    int l = 0, r = arr.size() - 1;
    while (l < r) {
      int mid = l + (r - l) / 2;
      if (sumRange(0, mid) < kth)
        l = mid + 1;
      else
        r = mid;
    }
    return l;
  }

 private:
  vector<int> arr;
  vector<int> bit_tree;
};
int main() {
  int n, idx = 1;
  cin >> n;
  vector<int> arr(n, 0);
  map<int, int> mp;
  for (auto& val : arr) {
    cin >> val;
    mp[val] = 0;
  }
  for (auto& p : mp) {
    p.second = idx++;
  }
  for (auto& val : arr) {
    val = mp[val];
  }
  vector<int> init(n + 2, 0);
  auto bit = new BinaryIndexTree(init);
  vector<long long> left(n, 0);
  vector<long long> right(n, 0);
  for (int i = 0; i < n; i++) {
    int sm_num = bit->sumRange(0, arr[i]);
    left[i] = (i)-sm_num;
    bit->update(arr[i], 1);
  }
  auto bit2 = new BinaryIndexTree(init);
  for (int i = n - 1; i >= 0; i--) {
    int sm_num = bit2->sumRange(0, arr[i]);
    right[i] = sm_num;
    bit2->update(arr[i], 1);
  }
  long long ans = 0;
  for (int i = 1; i < n - 1; i++) ans += left[i] * right[i];
  cout << ans << "\n";
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int static fast = []() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  return 0;
}();
class BinaryIndexTree {
 public:
  BinaryIndexTree(vector<int> nums) {
    int n = nums.size();
    arr = vector<int>(n, 0);
    bit_tree = vector<int>(n + 1, 0);
    for (int i = 0; i < n; i++) update(i, nums[i]);
  }
  void update(int idx, int val) {
    int delta = val - arr[idx];
    arr[idx] = val;
    int x = idx + 1;
    while (x < bit_tree.size()) {
      bit_tree[x] += delta;
      x += x & (-x);
    }
  }
  int _sumRange(int x) {
    int ret = 0;
    while (x > 0) {
      ret += bit_tree[x];
      x -= x & (-x);
    }
    return ret;
  }
  int sumRange(int from, int to) { return _sumRange(to + 1) - _sumRange(from); }
  int find_kth_smaller(int kth) {
    int l = 0, r = arr.size() - 1;
    while (l < r) {
      int mid = l + (r - l) / 2;
      if (sumRange(0, mid) < kth)
        l = mid + 1;
      else
        r = mid;
    }
    return l;
  }

 private:
  vector<int> arr;
  vector<int> bit_tree;
};
int main() {
  int n, idx = 1;
  cin >> n;
  vector<int> arr(n, 0);
  map<int, int> mp;
  for (auto& val : arr) {
    cin >> val;
    mp[val] = 0;
  }
  for (auto& p : mp) {
    p.second = idx++;
  }
  for (auto& val : arr) {
    val = mp[val];
  }
  vector<int> init(n + 2, 0);
  auto bit = new BinaryIndexTree(init);
  vector<long long> left(n, 0);
  vector<long long> right(n, 0);
  for (int i = 0; i < n; i++) {
    int sm_num = bit->sumRange(0, arr[i]);
    left[i] = (i)-sm_num;
    bit->update(arr[i], 1);
  }
  auto bit2 = new BinaryIndexTree(init);
  for (int i = n - 1; i >= 0; i--) {
    int sm_num = bit2->sumRange(0, arr[i]);
    right[i] = sm_num;
    bit2->update(arr[i], 1);
  }
  long long ans = 0;
  for (int i = 1; i < n - 1; i++) ans += left[i] * right[i];
  cout << ans << "\n";
  return 0;
}

```