#include <bits/stdc++.h>
using namespace std;
void MAIN();
int main() {
  MAIN();
  return 0;
}
const long long N = 1e6 + 1, mod = 1e9 + 7;
int n, m, dp[N][2];
pair<int, int> riki[N];
struct fwt {
  long long bit[N];
  void upd(int i, long long val) {
    for (; i < N; i += i & -i) {
      bit[i] += val;
    }
  }
  long long get(int i) {
    long long result = 0;
    for (; i > 0; i &= i - 1) {
      result += bit[i];
    }
    return result;
  }
} fwt1, fwt2, fwt3;
void MAIN() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int val;
    scanf("%d", &val);
    riki[i] = pair<int, int>(val, i);
  }
  sort(riki + 1, riki + n + 1);
  long long ans = 0;
  for (int i = n; i >= 1; --i) {
    ans += fwt2.get(riki[i].second);
    fwt2.upd(riki[i].second, fwt1.get(riki[i].second));
    fwt1.upd(riki[i].second, 1);
  }
  printf("%lld", ans);
}

### Prompt

Create a solution in Cpp for 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;
void MAIN();
int main() {
  MAIN();
  return 0;
}
const long long N = 1e6 + 1, mod = 1e9 + 7;
int n, m, dp[N][2];
pair<int, int> riki[N];
struct fwt {
  long long bit[N];
  void upd(int i, long long val) {
    for (; i < N; i += i & -i) {
      bit[i] += val;
    }
  }
  long long get(int i) {
    long long result = 0;
    for (; i > 0; i &= i - 1) {
      result += bit[i];
    }
    return result;
  }
} fwt1, fwt2, fwt3;
void MAIN() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int val;
    scanf("%d", &val);
    riki[i] = pair<int, int>(val, i);
  }
  sort(riki + 1, riki + n + 1);
  long long ans = 0;
  for (int i = n; i >= 1; --i) {
    ans += fwt2.get(riki[i].second);
    fwt2.upd(riki[i].second, fwt1.get(riki[i].second));
    fwt1.upd(riki[i].second, 1);
  }
  printf("%lld", ans);
}

```