#include <bits/stdc++.h>
using namespace std;
struct Node {
  int pos;
  int val;
} recode[1000010];
long long ad[1000010], lb[1000010], rs[1000010], ls[1000010], n, c[1000010];
long long ans;
int cmp(Node x, Node y) { return x.val < y.val; }
int lowbit(int x) { return x & (-x); }
void add(int po, int val) {
  while (po <= n + 1) {
    c[po] += val;
    po += lowbit(po);
  }
}
int getsum(int po) {
  long long sum = 0;
  while (po > 0) {
    sum += c[po];
    po -= lowbit(po);
  }
  return sum;
}
int main() {
  scanf("%d", &n);
  ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &recode[i].val);
    recode[i].pos = i;
  }
  sort(recode + 1, recode + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    ad[recode[i].pos] = i;
  }
  for (int i = 1; i <= n; i++) {
    add(ad[i], 1);
    ls[ad[i]] = getsum(ad[i]) - 1;
    lb[ad[i]] = i - ls[ad[i]] - 1;
    rs[ad[i]] = ad[i] - ls[ad[i]] - 1;
    ans += lb[ad[i]] * rs[ad[i]];
  }
  printf("%I64d\n", ans);
}

### 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;
struct Node {
  int pos;
  int val;
} recode[1000010];
long long ad[1000010], lb[1000010], rs[1000010], ls[1000010], n, c[1000010];
long long ans;
int cmp(Node x, Node y) { return x.val < y.val; }
int lowbit(int x) { return x & (-x); }
void add(int po, int val) {
  while (po <= n + 1) {
    c[po] += val;
    po += lowbit(po);
  }
}
int getsum(int po) {
  long long sum = 0;
  while (po > 0) {
    sum += c[po];
    po -= lowbit(po);
  }
  return sum;
}
int main() {
  scanf("%d", &n);
  ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &recode[i].val);
    recode[i].pos = i;
  }
  sort(recode + 1, recode + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    ad[recode[i].pos] = i;
  }
  for (int i = 1; i <= n; i++) {
    add(ad[i], 1);
    ls[ad[i]] = getsum(ad[i]) - 1;
    lb[ad[i]] = i - ls[ad[i]] - 1;
    rs[ad[i]] = ad[i] - ls[ad[i]] - 1;
    ans += lb[ad[i]] * rs[ad[i]];
  }
  printf("%I64d\n", ans);
}

```