#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, a[N], b[N], rgt[4 * N], lft[4 * N];
map<int, int> id;
void compress() {
  int k = 0;
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    id[b[i]] = ++k;
  }
}
void update(int p, int l, int r, int i, int v, int k) {
  if (l > i || r < i) return;
  if (l >= i && r <= i) {
    if (k)
      rgt[p] += v;
    else
      lft[p] += v;
    return;
  }
  int m = (l + r) >> 1;
  update(p * 2, l, m, i, v, k);
  update(p * 2 + 1, m + 1, r, i, v, k);
  rgt[p] = rgt[p * 2] + rgt[p * 2 + 1];
  lft[p] = lft[p * 2] + lft[p * 2 + 1];
}
int query(int p, int l, int r, int i, int j, int k) {
  if (i > r || j < l) return 0;
  if (l >= i && r <= j) {
    if (k)
      return rgt[p];
    else
      return lft[p];
  }
  int m = (l + r) >> 1;
  return query(p * 2, l, m, i, j, k) + query(p * 2 + 1, m + 1, r, i, j, k);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    b[i] = a[i];
  }
  compress();
  for (int i = 1; i < n; i++) {
    update(1, 1, n, id[a[i]], 1, 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long x = query(1, 1, n, 1, id[a[i]] - 1, 1);
    long long y = query(1, 1, n, id[a[i]] + 1, n, 0);
    if (i > 0) update(1, 1, n, id[a[i]], -1, 1);
    update(1, 1, n, id[a[i]], 1, 0);
    ans += (x * y);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Your task is to create 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 + 5;
int n, a[N], b[N], rgt[4 * N], lft[4 * N];
map<int, int> id;
void compress() {
  int k = 0;
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    id[b[i]] = ++k;
  }
}
void update(int p, int l, int r, int i, int v, int k) {
  if (l > i || r < i) return;
  if (l >= i && r <= i) {
    if (k)
      rgt[p] += v;
    else
      lft[p] += v;
    return;
  }
  int m = (l + r) >> 1;
  update(p * 2, l, m, i, v, k);
  update(p * 2 + 1, m + 1, r, i, v, k);
  rgt[p] = rgt[p * 2] + rgt[p * 2 + 1];
  lft[p] = lft[p * 2] + lft[p * 2 + 1];
}
int query(int p, int l, int r, int i, int j, int k) {
  if (i > r || j < l) return 0;
  if (l >= i && r <= j) {
    if (k)
      return rgt[p];
    else
      return lft[p];
  }
  int m = (l + r) >> 1;
  return query(p * 2, l, m, i, j, k) + query(p * 2 + 1, m + 1, r, i, j, k);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    b[i] = a[i];
  }
  compress();
  for (int i = 1; i < n; i++) {
    update(1, 1, n, id[a[i]], 1, 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long x = query(1, 1, n, 1, id[a[i]] - 1, 1);
    long long y = query(1, 1, n, id[a[i]] + 1, n, 0);
    if (i > 0) update(1, 1, n, id[a[i]], -1, 1);
    update(1, 1, n, id[a[i]], 1, 0);
    ans += (x * y);
  }
  cout << ans << '\n';
  return 0;
}

```