#include <bits/stdc++.h>
using namespace std;
const int SZ = 1000010;
const int INF = 1e9 + 10;
const int mod = 1e9 + 7;
const long double eps = 1e-12;
long long read() {
  long long n = 0;
  char a = getchar();
  bool flag = 0;
  while (a > '9' || a < '0') {
    if (a == '-') flag = 1;
    a = getchar();
  }
  while (a <= '9' && a >= '0') {
    n = n * 10 + a - '0', a = getchar();
  }
  if (flag) n = -n;
  return n;
}
int tree[SZ], L[SZ], R[SZ], n, lsh[SZ], a[SZ];
void add(int x, int d) {
  for (int i = x; i <= n; i += i & -i) tree[i] += d;
}
int ask(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= i & -i) ans += tree[i];
  return ans;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) lsh[i] = a[i] = read();
  sort(lsh + 1, lsh + 1 + n);
  int m = unique(lsh + 1, lsh + 1 + n) - lsh - 1;
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(lsh + 1, lsh + 1 + n, a[i]) - lsh;
  for (int i = 1; i <= n; i++) {
    L[i] = i - 1 - ask(a[i]);
    add(a[i], 1);
  }
  for (int i = 1; i <= n; i++) tree[i] = 0;
  long long ans = 0;
  for (int i = n; i >= 1; i--) {
    R[i] = ask(a[i]);
    add(a[i], 1);
    ans += 1ll * R[i] * L[i];
  }
  cout << ans << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 SZ = 1000010;
const int INF = 1e9 + 10;
const int mod = 1e9 + 7;
const long double eps = 1e-12;
long long read() {
  long long n = 0;
  char a = getchar();
  bool flag = 0;
  while (a > '9' || a < '0') {
    if (a == '-') flag = 1;
    a = getchar();
  }
  while (a <= '9' && a >= '0') {
    n = n * 10 + a - '0', a = getchar();
  }
  if (flag) n = -n;
  return n;
}
int tree[SZ], L[SZ], R[SZ], n, lsh[SZ], a[SZ];
void add(int x, int d) {
  for (int i = x; i <= n; i += i & -i) tree[i] += d;
}
int ask(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= i & -i) ans += tree[i];
  return ans;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) lsh[i] = a[i] = read();
  sort(lsh + 1, lsh + 1 + n);
  int m = unique(lsh + 1, lsh + 1 + n) - lsh - 1;
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(lsh + 1, lsh + 1 + n, a[i]) - lsh;
  for (int i = 1; i <= n; i++) {
    L[i] = i - 1 - ask(a[i]);
    add(a[i], 1);
  }
  for (int i = 1; i <= n; i++) tree[i] = 0;
  long long ans = 0;
  for (int i = n; i >= 1; i--) {
    R[i] = ask(a[i]);
    add(a[i], 1);
    ans += 1ll * R[i] * L[i];
  }
  cout << ans << endl;
}

```