#include <bits/stdc++.h>
using namespace std;
const int MaxN = 1000 * 1000;
pair<int, int> a[MaxN];
int b[MaxN];
int t[(MaxN + 2) * 4];
int sum(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (l == tl && r == tr) return t[v];
  int mid = (tl + tr) / 2;
  return sum(v * 2, tl, mid, l, min(r, mid)) +
         sum(v * 2 + 1, mid + 1, tr, max(l, mid + 1), r);
}
void add(int v, int tl, int tr, int pos, int val) {
  if (tl == tr)
    t[v] += val;
  else {
    int mid = (tl + tr) / 2;
    if (pos <= mid)
      add(v * 2, tl, mid, pos, val);
    else
      add(v * 2 + 1, mid + 1, tr, pos, val);
    t[v] = t[v * 2] + t[v * 2 + 1];
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) b[a[i].second] = i + 1;
  long long res = 0;
  for (int i = 0; i < n; i++) {
    add(1, 0, n + 1, b[i], 1);
    res += sum(1, 0, n + 1, b[i] + 1, n + 1) * 1ll *
           (b[i] - 1 - sum(1, 0, n + 1, 0, b[i] - 1));
  }
  printf("%lld", res);
}

### 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;
const int MaxN = 1000 * 1000;
pair<int, int> a[MaxN];
int b[MaxN];
int t[(MaxN + 2) * 4];
int sum(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (l == tl && r == tr) return t[v];
  int mid = (tl + tr) / 2;
  return sum(v * 2, tl, mid, l, min(r, mid)) +
         sum(v * 2 + 1, mid + 1, tr, max(l, mid + 1), r);
}
void add(int v, int tl, int tr, int pos, int val) {
  if (tl == tr)
    t[v] += val;
  else {
    int mid = (tl + tr) / 2;
    if (pos <= mid)
      add(v * 2, tl, mid, pos, val);
    else
      add(v * 2 + 1, mid + 1, tr, pos, val);
    t[v] = t[v * 2] + t[v * 2 + 1];
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) b[a[i].second] = i + 1;
  long long res = 0;
  for (int i = 0; i < n; i++) {
    add(1, 0, n + 1, b[i], 1);
    res += sum(1, 0, n + 1, b[i] + 1, n + 1) * 1ll *
           (b[i] - 1 - sum(1, 0, n + 1, 0, b[i] - 1));
  }
  printf("%lld", res);
}

```