#include <bits/stdc++.h>
using namespace std;
int tree[1000000 * 4], tree1[1000000 * 4];
pair<int, int> a[1000003];
int n;
long long jaw = 0;
void upd(int idx, int l, int r, int in, int val) {
  if (in > r || in < l) return;
  if (l == r) {
    tree[idx] += val;
    return;
  }
  int mid = (l + r) / 2;
  upd(idx * 2, l, mid, in, val);
  upd(idx * 2 + 1, mid + 1, r, in, val);
  tree[idx] = tree[idx * 2 + 1] + tree[idx * 2];
  return;
}
void upd1(int idx, int l, int r, int in, int val) {
  if (in > r || in < l) return;
  if (l == r) {
    tree1[idx] += val;
    return;
  }
  int mid = (l + r) / 2;
  upd1(idx * 2, l, mid, in, val);
  upd1(idx * 2 + 1, mid + 1, r, in, val);
  tree1[idx] = tree1[idx * 2 + 1] + tree1[idx * 2];
  return;
}
int cari(int idx, int l, int r, int fl, int fr) {
  if (l > fr || fl > r) return 0;
  if (fl <= l && r <= fr) return tree[idx];
  int mid = (l + r) / 2;
  return cari(idx * 2, l, mid, fl, fr) + cari(idx * 2 + 1, mid + 1, r, fl, fr);
}
int cari1(int idx, int l, int r, int fl, int fr) {
  if (l > fr || fl > r) return 0;
  if (fl <= l && r <= fr) return tree1[idx];
  int mid = (l + r) / 2;
  return cari1(idx * 2, l, mid, fl, fr) +
         cari1(idx * 2 + 1, mid + 1, r, fl, fr);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    upd(1, 1, n, a[i].second, 1);
  }
  for (int i = n; i > 0; i--) {
    upd(1, 1, n, a[i].second, -1);
    int sa = cari(1, 1, n, a[i].second + 1, n);
    int du = cari1(1, 1, n, 1, a[i].second - 1);
    upd1(1, 1, n, a[i].second, 1);
    jaw += (long long)sa * du;
  }
  cout << jaw << endl;
}

### 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;
int tree[1000000 * 4], tree1[1000000 * 4];
pair<int, int> a[1000003];
int n;
long long jaw = 0;
void upd(int idx, int l, int r, int in, int val) {
  if (in > r || in < l) return;
  if (l == r) {
    tree[idx] += val;
    return;
  }
  int mid = (l + r) / 2;
  upd(idx * 2, l, mid, in, val);
  upd(idx * 2 + 1, mid + 1, r, in, val);
  tree[idx] = tree[idx * 2 + 1] + tree[idx * 2];
  return;
}
void upd1(int idx, int l, int r, int in, int val) {
  if (in > r || in < l) return;
  if (l == r) {
    tree1[idx] += val;
    return;
  }
  int mid = (l + r) / 2;
  upd1(idx * 2, l, mid, in, val);
  upd1(idx * 2 + 1, mid + 1, r, in, val);
  tree1[idx] = tree1[idx * 2 + 1] + tree1[idx * 2];
  return;
}
int cari(int idx, int l, int r, int fl, int fr) {
  if (l > fr || fl > r) return 0;
  if (fl <= l && r <= fr) return tree[idx];
  int mid = (l + r) / 2;
  return cari(idx * 2, l, mid, fl, fr) + cari(idx * 2 + 1, mid + 1, r, fl, fr);
}
int cari1(int idx, int l, int r, int fl, int fr) {
  if (l > fr || fl > r) return 0;
  if (fl <= l && r <= fr) return tree1[idx];
  int mid = (l + r) / 2;
  return cari1(idx * 2, l, mid, fl, fr) +
         cari1(idx * 2 + 1, mid + 1, r, fl, fr);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    upd(1, 1, n, a[i].second, 1);
  }
  for (int i = n; i > 0; i--) {
    upd(1, 1, n, a[i].second, -1);
    int sa = cari(1, 1, n, a[i].second + 1, n);
    int du = cari1(1, 1, n, 1, a[i].second - 1);
    upd1(1, 1, n, a[i].second, 1);
    jaw += (long long)sa * du;
  }
  cout << jaw << endl;
}

```