#include <bits/stdc++.h>
using namespace std;
const int N = 1000001;
long long int tree[N * 3];
long long int tree2[N * 3];
long long int dp[N];
int ar[N];
map<int, int> cor;
vector<int> lis;
void update(long long int* tr, int k, int l, int r, int p, long long int v) {
  if (l == r) {
    if (l == p) {
      tr[k] += v;
    }
    return;
  }
  if (l > p || r < p) {
    return;
  }
  update(tr, k * 2, l, (l + r) / 2, p, v);
  update(tr, k * 2 + 1, (l + r) / 2 + 1, r, p, v);
  tr[k] = tr[k * 2] + tr[k * 2 + 1];
}
long long int query(long long int* tr, int k, int l, int r, int ql, int qr) {
  if (ql > qr) {
    return 0;
  }
  if (ql <= l && r <= qr) {
    return tr[k];
  }
  if (l > qr || r < ql) {
    return 0;
  }
  return query(tr, k * 2, l, (l + r) / 2, ql, qr) +
         query(tr, k * 2 + 1, (l + r) / 2 + 1, r, ql, qr);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &ar[i]);
    if (cor[ar[i]] == 0) {
      lis.push_back(ar[i]);
      cor[ar[i]] = 1;
    }
  }
  cor.clear();
  int id = 0;
  sort(lis.begin(), lis.end());
  for (int i = 0; i < lis.size(); i++) {
    cor[lis[i]] = id++;
  }
  long long int ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    ans += query(tree2, 1, 0, n - 1, 0, cor[ar[i]] - 1);
    update(tree, 1, 0, n - 1, cor[ar[i]], 1);
    update(tree2, 1, 0, n - 1, cor[ar[i]],
           query(tree, 1, 0, n - 1, 0, cor[ar[i]] - 1));
  }
  printf("%lld\n", 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;
const int N = 1000001;
long long int tree[N * 3];
long long int tree2[N * 3];
long long int dp[N];
int ar[N];
map<int, int> cor;
vector<int> lis;
void update(long long int* tr, int k, int l, int r, int p, long long int v) {
  if (l == r) {
    if (l == p) {
      tr[k] += v;
    }
    return;
  }
  if (l > p || r < p) {
    return;
  }
  update(tr, k * 2, l, (l + r) / 2, p, v);
  update(tr, k * 2 + 1, (l + r) / 2 + 1, r, p, v);
  tr[k] = tr[k * 2] + tr[k * 2 + 1];
}
long long int query(long long int* tr, int k, int l, int r, int ql, int qr) {
  if (ql > qr) {
    return 0;
  }
  if (ql <= l && r <= qr) {
    return tr[k];
  }
  if (l > qr || r < ql) {
    return 0;
  }
  return query(tr, k * 2, l, (l + r) / 2, ql, qr) +
         query(tr, k * 2 + 1, (l + r) / 2 + 1, r, ql, qr);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &ar[i]);
    if (cor[ar[i]] == 0) {
      lis.push_back(ar[i]);
      cor[ar[i]] = 1;
    }
  }
  cor.clear();
  int id = 0;
  sort(lis.begin(), lis.end());
  for (int i = 0; i < lis.size(); i++) {
    cor[lis[i]] = id++;
  }
  long long int ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    ans += query(tree2, 1, 0, n - 1, 0, cor[ar[i]] - 1);
    update(tree, 1, 0, n - 1, cor[ar[i]], 1);
    update(tree2, 1, 0, n - 1, cor[ar[i]],
           query(tree, 1, 0, n - 1, 0, cor[ar[i]] - 1));
  }
  printf("%lld\n", ans);
}

```