#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const long long N = 1000004;
long long tree[N];
long long ara[N], temp[N];
unordered_map<long long, long long> mp;
long long n;
long long read_int() {
  bool minus = false;
  long long result = 0;
  char ch;
  ch = getchar();
  while (true) {
    if (ch == '-') break;
    if (ch >= '0' && ch <= '9') break;
    ch = getchar();
  }
  if (ch == '-')
    minus = true;
  else
    result = ch - '0';
  while (true) {
    ch = getchar();
    if (ch < '0' || ch > '9') break;
    result = result * 10 + (ch - '0');
  }
  if (minus)
    return -result;
  else
    return result;
}
ll get(long long indx) {
  ll sum = 0;
  while (indx > 0) {
    sum += tree[indx];
    indx -= indx & -indx;
  }
  return sum;
}
void update(long long indx, ll val) {
  while (indx <= n) {
    tree[indx] += val;
    indx += indx & -indx;
  }
}
int32_t main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> ara[i];
    temp[i] = ara[i];
  }
  sort(temp + 1, temp + 1 + n);
  for (long long i = 1; i <= n; i++)
    if (!mp[temp[i]]) mp[temp[i]] = i;
  for (long long i = 1; i <= n; i++) ara[i] = mp[ara[i]];
  for (long long i = n; i >= 1; i--) {
    ll k = get(ara[i] - 1);
    temp[i] = k;
    update(ara[i], 1);
  }
  long long sum = 0;
  memset(tree, 0, sizeof tree);
  for (long long i = n; i >= 1; i--) {
    long long k = get(ara[i]);
    sum += k;
    update(ara[i], temp[i]);
  }
  printf("%lld\n", sum);
  return 0;
}

### Prompt

Please create a solution in cpp 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;
using ll = long long;
const long long N = 1000004;
long long tree[N];
long long ara[N], temp[N];
unordered_map<long long, long long> mp;
long long n;
long long read_int() {
  bool minus = false;
  long long result = 0;
  char ch;
  ch = getchar();
  while (true) {
    if (ch == '-') break;
    if (ch >= '0' && ch <= '9') break;
    ch = getchar();
  }
  if (ch == '-')
    minus = true;
  else
    result = ch - '0';
  while (true) {
    ch = getchar();
    if (ch < '0' || ch > '9') break;
    result = result * 10 + (ch - '0');
  }
  if (minus)
    return -result;
  else
    return result;
}
ll get(long long indx) {
  ll sum = 0;
  while (indx > 0) {
    sum += tree[indx];
    indx -= indx & -indx;
  }
  return sum;
}
void update(long long indx, ll val) {
  while (indx <= n) {
    tree[indx] += val;
    indx += indx & -indx;
  }
}
int32_t main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> ara[i];
    temp[i] = ara[i];
  }
  sort(temp + 1, temp + 1 + n);
  for (long long i = 1; i <= n; i++)
    if (!mp[temp[i]]) mp[temp[i]] = i;
  for (long long i = 1; i <= n; i++) ara[i] = mp[ara[i]];
  for (long long i = n; i >= 1; i--) {
    ll k = get(ara[i] - 1);
    temp[i] = k;
    update(ara[i], 1);
  }
  long long sum = 0;
  memset(tree, 0, sizeof tree);
  for (long long i = n; i >= 1; i--) {
    long long k = get(ara[i]);
    sum += k;
    update(ara[i], temp[i]);
  }
  printf("%lld\n", sum);
  return 0;
}

```