#include <bits/stdc++.h>
using namespace std;
long long BIT1[1000010], BIT2[1000010], arr[1000010], val[1000010];
map<long long, long long> m;
long long ans[1000010];
void pointupdate(long long index, long long val, long long n, long long BIT[]) {
  for (long long i = index; i <= n; i += (i & (-i))) {
    BIT[i] += val;
  }
}
long long getsum(long long index, long long n, long long BIT[]) {
  long long sum = 0;
  for (long long i = index; i > 0; i -= (i & (-i))) {
    sum += BIT[i];
  }
  return sum;
}
int main() {
  long long n;
  cin >> n;
  long long k = 0;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    if (!m[arr[i]]) {
      val[k++] = arr[i];
      m[arr[i]]++;
    }
  }
  sort(val, val + k);
  for (long long i = 0; i < k; i++) {
    m[val[i]] = i + 1;
  }
  long long k1 = 0;
  for (long long i = 0; i < n; i++) {
    long long x = m[arr[i]];
    if (x == k)
      ans[i] = 0;
    else {
      long long sum1 = getsum(k, k, BIT1);
      long long sum2 = getsum(x, k, BIT1);
      ans[i] = sum1 - sum2;
    }
    pointupdate(x, 1, n, BIT1);
  }
  long long ans1 = 0;
  for (long long i = n - 1; i >= 0; i--) {
    long long x = m[arr[i]];
    if (x > 1) {
      long long sum1 = getsum(x - 1, k, BIT2);
      if (sum1) ans1 = ans1 + ans[i] * sum1;
    }
    pointupdate(x, 1, k, BIT2);
  }
  cout << ans1 << endl;
}

### 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;
long long BIT1[1000010], BIT2[1000010], arr[1000010], val[1000010];
map<long long, long long> m;
long long ans[1000010];
void pointupdate(long long index, long long val, long long n, long long BIT[]) {
  for (long long i = index; i <= n; i += (i & (-i))) {
    BIT[i] += val;
  }
}
long long getsum(long long index, long long n, long long BIT[]) {
  long long sum = 0;
  for (long long i = index; i > 0; i -= (i & (-i))) {
    sum += BIT[i];
  }
  return sum;
}
int main() {
  long long n;
  cin >> n;
  long long k = 0;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    if (!m[arr[i]]) {
      val[k++] = arr[i];
      m[arr[i]]++;
    }
  }
  sort(val, val + k);
  for (long long i = 0; i < k; i++) {
    m[val[i]] = i + 1;
  }
  long long k1 = 0;
  for (long long i = 0; i < n; i++) {
    long long x = m[arr[i]];
    if (x == k)
      ans[i] = 0;
    else {
      long long sum1 = getsum(k, k, BIT1);
      long long sum2 = getsum(x, k, BIT1);
      ans[i] = sum1 - sum2;
    }
    pointupdate(x, 1, n, BIT1);
  }
  long long ans1 = 0;
  for (long long i = n - 1; i >= 0; i--) {
    long long x = m[arr[i]];
    if (x > 1) {
      long long sum1 = getsum(x - 1, k, BIT2);
      if (sum1) ans1 = ans1 + ans[i] * sum1;
    }
    pointupdate(x, 1, k, BIT2);
  }
  cout << ans1 << endl;
}

```