#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 7;
struct node {
  long long data, id;
} arr[maxn];
long long c[maxn], n;
long long l[maxn], r[maxn];
long long lowBit(long long x) { return x & -x; }
void treeAdd(long long pos, long long val) {
  for (long long i = pos; i <= n; i += lowBit(i)) {
    c[i] += val;
  }
}
long long treeSum(long long pos) {
  long long sum = 0;
  for (long long i = pos; i > 0; i -= lowBit(i)) {
    sum += c[i];
  }
  return sum;
}
bool cmp(node a, node b) { return a.data > b.data; }
int main() {
  while (~scanf("%I64d", &n)) {
    for (long long i = 1; i <= n; i++) {
      scanf("%I64d", &arr[i].data);
      arr[i].id = i;
    }
    sort(arr + 1, arr + 1 + n, cmp);
    memset(c, 0, sizeof(c));
    for (long long i = 1; i <= n; i++) {
      l[i] = treeSum(arr[i].id);
      treeAdd(arr[i].id, 1);
    }
    memset(c, 0, sizeof(c));
    for (long long i = n; i >= 1; i--) {
      r[i] = treeSum(n) - treeSum(arr[i].id - 1);
      treeAdd(arr[i].id, 1);
    }
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      ans += l[i] * r[i];
    }
    printf("%I64d\n", ans);
  }
  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 maxn = 1e6 + 7;
struct node {
  long long data, id;
} arr[maxn];
long long c[maxn], n;
long long l[maxn], r[maxn];
long long lowBit(long long x) { return x & -x; }
void treeAdd(long long pos, long long val) {
  for (long long i = pos; i <= n; i += lowBit(i)) {
    c[i] += val;
  }
}
long long treeSum(long long pos) {
  long long sum = 0;
  for (long long i = pos; i > 0; i -= lowBit(i)) {
    sum += c[i];
  }
  return sum;
}
bool cmp(node a, node b) { return a.data > b.data; }
int main() {
  while (~scanf("%I64d", &n)) {
    for (long long i = 1; i <= n; i++) {
      scanf("%I64d", &arr[i].data);
      arr[i].id = i;
    }
    sort(arr + 1, arr + 1 + n, cmp);
    memset(c, 0, sizeof(c));
    for (long long i = 1; i <= n; i++) {
      l[i] = treeSum(arr[i].id);
      treeAdd(arr[i].id, 1);
    }
    memset(c, 0, sizeof(c));
    for (long long i = n; i >= 1; i--) {
      r[i] = treeSum(n) - treeSum(arr[i].id - 1);
      treeAdd(arr[i].id, 1);
    }
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      ans += l[i] * r[i];
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```