#include <bits/stdc++.h>
using namespace std;
const int N = (1e6) + 5;
struct Soldier {
  int value;
  int index;
};
Soldier a[N];
bool Compare(Soldier p, Soldier q) { return p.value < q.value; }
int power[N], tree[N];
long long l[N], r[N];
int n;
int Query(int ind) {
  int sum = 0;
  while (ind >= 1) {
    sum += tree[ind];
    ind -= (ind & (-ind));
  }
  return sum;
}
int GetQuery(int low, int high) { return Query(high) - Query(low - 1); }
void Update(int ind, int delta) {
  while (ind <= n) {
    tree[ind] += delta;
    ind += (ind & (-ind));
  }
}
int main() {
  int i;
  long long ans;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i].value);
    a[i].index = i;
  }
  sort(a + 1, a + n + 1, Compare);
  for (i = 1; i <= n; i++) power[a[i].index] = i;
  for (i = 1; i <= n; i++) {
    l[i] = GetQuery(power[i] + 1, n);
    Update(power[i], 1);
  }
  for (i = 1; i <= n; i++) r[i] = (power[i] - 1) - (i - 1 - l[i]);
  ans = 0;
  for (i = 1; i <= n; i++) ans += l[i] * r[i];
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
struct Soldier {
  int value;
  int index;
};
Soldier a[N];
bool Compare(Soldier p, Soldier q) { return p.value < q.value; }
int power[N], tree[N];
long long l[N], r[N];
int n;
int Query(int ind) {
  int sum = 0;
  while (ind >= 1) {
    sum += tree[ind];
    ind -= (ind & (-ind));
  }
  return sum;
}
int GetQuery(int low, int high) { return Query(high) - Query(low - 1); }
void Update(int ind, int delta) {
  while (ind <= n) {
    tree[ind] += delta;
    ind += (ind & (-ind));
  }
}
int main() {
  int i;
  long long ans;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i].value);
    a[i].index = i;
  }
  sort(a + 1, a + n + 1, Compare);
  for (i = 1; i <= n; i++) power[a[i].index] = i;
  for (i = 1; i <= n; i++) {
    l[i] = GetQuery(power[i] + 1, n);
    Update(power[i], 1);
  }
  for (i = 1; i <= n; i++) r[i] = (power[i] - 1) - (i - 1 - l[i]);
  ans = 0;
  for (i = 1; i <= n; i++) ans += l[i] * r[i];
  printf("%lld\n", ans);
  return 0;
}

```