#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
inline int getint() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long getlng() {
  long long x;
  scanf("%lld", &x);
  return x;
}
int dx4[] = {-1, 0, 1, 0};
int dy4[] = {0, 1, 0, -1};
int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const int N = 1000001;
int arr[N];
vector<int> tree[N << 2];
long long ans = 0;
void build(int node, int b, int e) {
  if (b == e) {
    tree[node].push_back(arr[b]);
    return;
  }
  int lft = node << 1, rgt = lft | 1, mid = (b + e) >> 1;
  ;
  build(lft, b, mid);
  build(rgt, mid + 1, e);
  merge(tree[lft].begin(), tree[lft].end(), tree[rgt].begin(), tree[rgt].end(),
        back_inserter(tree[node]));
}
int Q(int node, int b, int e, int i, int j, int k) {
  if (b > j || e < i) return 0;
  if (b >= i && e <= j)
    return lower_bound(tree[node].begin(), tree[node].end(), k) -
           tree[node].begin();
  int lft = node << 1, rgt = lft | 1, mid = (b + e) >> 1;
  ;
  return Q(lft, b, mid, i, j, k) + Q(rgt, mid + 1, e, i, j, k);
}
int main() {
  int n = getint();
  for (int i = 1; i <= n; ++i) arr[i] = getint();
  build(1, 1, n);
  for (int i = 2; i < n; ++i) {
    long long p = (long long)(Q(1, 1, n, 1, i - 1, arr[i]));
    p = i - 1 - p;
    long long q = (long long)(Q(1, 1, n, i + 1, n, arr[i]));
    ans += p * q;
  }
  printf("%lld\n", ans);
  return 0;
}

### 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 double PI = acos(-1.0);
inline int getint() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long getlng() {
  long long x;
  scanf("%lld", &x);
  return x;
}
int dx4[] = {-1, 0, 1, 0};
int dy4[] = {0, 1, 0, -1};
int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const int N = 1000001;
int arr[N];
vector<int> tree[N << 2];
long long ans = 0;
void build(int node, int b, int e) {
  if (b == e) {
    tree[node].push_back(arr[b]);
    return;
  }
  int lft = node << 1, rgt = lft | 1, mid = (b + e) >> 1;
  ;
  build(lft, b, mid);
  build(rgt, mid + 1, e);
  merge(tree[lft].begin(), tree[lft].end(), tree[rgt].begin(), tree[rgt].end(),
        back_inserter(tree[node]));
}
int Q(int node, int b, int e, int i, int j, int k) {
  if (b > j || e < i) return 0;
  if (b >= i && e <= j)
    return lower_bound(tree[node].begin(), tree[node].end(), k) -
           tree[node].begin();
  int lft = node << 1, rgt = lft | 1, mid = (b + e) >> 1;
  ;
  return Q(lft, b, mid, i, j, k) + Q(rgt, mid + 1, e, i, j, k);
}
int main() {
  int n = getint();
  for (int i = 1; i <= n; ++i) arr[i] = getint();
  build(1, 1, n);
  for (int i = 2; i < n; ++i) {
    long long p = (long long)(Q(1, 1, n, 1, i - 1, arr[i]));
    p = i - 1 - p;
    long long q = (long long)(Q(1, 1, n, i + 1, n, arr[i]));
    ans += p * q;
  }
  printf("%lld\n", ans);
  return 0;
}

```