#include <bits/stdc++.h>
using namespace std;
int i, j, k, l, n, m, T[1000005], x1, x2;
long long ans[1000005], Ans;
struct Nod {
  int x, y;
} a[1000005], b[1000005];
inline int lowbit(int x) { return (x & (-x)); }
inline void Up(int x) {
  for (int i = x; i <= 1000000; i += lowbit(i)) T[i]++;
}
inline int Ask(int x) {
  int tmp = 0;
  for (; x; x -= lowbit(x)) tmp += T[x];
  return tmp;
}
inline bool cmp(Nod l, Nod r) { return l.x < r.x; }
int main() {
  for (; ~scanf("%d", &n);) {
    for (i = 0; i <= 1000000; i++) T[i] = ans[i] = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i].x);
      a[i].y = i;
    }
    for (i = 1; i <= n; i++) b[i] = a[i];
    sort(b + 1, b + n + 1, cmp);
    for (i = 1; i <= n; i++) a[i].y = b[i].y;
    for (i = 1; i <= n; i++) {
      x1 = Ask(1000000);
      x2 = Ask(a[i].y);
      ans[i] = (long long)x1 - (long long)x2;
      Up(a[i].y);
    }
    for (i = 0; i <= 1000000; i++) T[i] = 0;
    for (i = n; i; i--) {
      ans[i] *= (long long)Ask(a[i].y);
      Up(a[i].y);
    }
    Ans = 0;
    for (i = 1; i <= n; i++) Ans += ans[i];
    printf("%lld\n", Ans);
  }
}

### Prompt

Your challenge is to write 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;
int i, j, k, l, n, m, T[1000005], x1, x2;
long long ans[1000005], Ans;
struct Nod {
  int x, y;
} a[1000005], b[1000005];
inline int lowbit(int x) { return (x & (-x)); }
inline void Up(int x) {
  for (int i = x; i <= 1000000; i += lowbit(i)) T[i]++;
}
inline int Ask(int x) {
  int tmp = 0;
  for (; x; x -= lowbit(x)) tmp += T[x];
  return tmp;
}
inline bool cmp(Nod l, Nod r) { return l.x < r.x; }
int main() {
  for (; ~scanf("%d", &n);) {
    for (i = 0; i <= 1000000; i++) T[i] = ans[i] = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i].x);
      a[i].y = i;
    }
    for (i = 1; i <= n; i++) b[i] = a[i];
    sort(b + 1, b + n + 1, cmp);
    for (i = 1; i <= n; i++) a[i].y = b[i].y;
    for (i = 1; i <= n; i++) {
      x1 = Ask(1000000);
      x2 = Ask(a[i].y);
      ans[i] = (long long)x1 - (long long)x2;
      Up(a[i].y);
    }
    for (i = 0; i <= 1000000; i++) T[i] = 0;
    for (i = n; i; i--) {
      ans[i] *= (long long)Ask(a[i].y);
      Up(a[i].y);
    }
    Ans = 0;
    for (i = 1; i <= n; i++) Ans += ans[i];
    printf("%lld\n", Ans);
  }
}

```