#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 1e6 + 6;
int n;
vector<int> a;
struct Fen {
  ll* bit;
  int s;
  vector<int> cols;
  map<int, int> colToIdx;
  void init(int _s) {
    s = _s;
    bit = new ll[s + 1];
    for (int i = 0; i <= s; i++) bit[i] = 0;
  }
  void init(const vector<int>& _c) {
    s = _c.size();
    cols = _c;
    sort(begin(cols), end(cols));
    cols.erase(unique(begin(cols), end(cols)), cols.end());
    bit = new ll[s + 1];
    for (int i = 0; i <= s; i++) bit[i] = 0;
    colToIdx[0] = 0;
    for (int i = 0; i < cols.size(); i++) {
      colToIdx[cols[i]] = i + 1;
    }
  }
  void add(int i, int x) {
    for (; i <= s; i += i & (-i)) bit[i] = (bit[i] + x);
  }
  ll sum(int i) {
    if (i == 0) return 0;
    ll r = 0;
    for (; i; i -= i & (-i)) {
      r = (r + bit[i]);
    }
    return r;
  }
  int idxForQuery(int i) {
    auto it = upper_bound(begin(cols), end(cols), i);
    if (it == cols.begin()) return 0;
    --it;
    return colToIdx[*it];
  }
  void insertValue(int col, int x) {
    int i = colToIdx[col];
    add(i, x);
  }
  ll getValue(int col) {
    int i = idxForQuery(col);
    return sum(i);
  }
};
int main() {
  scanf("%d", &n);
  a.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  Fen inv, cnt;
  inv.init(a);
  cnt.init(a);
  ll ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    ans += cnt.getValue(a[i] - 1);
    int c = inv.getValue(a[i] - 1);
    inv.insertValue(a[i], 1);
    cnt.insertValue(a[i], c);
  }
  cout << ans << '\n';
}

### Prompt

Develop a solution in cpp 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;
using ll = long long;
const int maxn = 1e6 + 6;
int n;
vector<int> a;
struct Fen {
  ll* bit;
  int s;
  vector<int> cols;
  map<int, int> colToIdx;
  void init(int _s) {
    s = _s;
    bit = new ll[s + 1];
    for (int i = 0; i <= s; i++) bit[i] = 0;
  }
  void init(const vector<int>& _c) {
    s = _c.size();
    cols = _c;
    sort(begin(cols), end(cols));
    cols.erase(unique(begin(cols), end(cols)), cols.end());
    bit = new ll[s + 1];
    for (int i = 0; i <= s; i++) bit[i] = 0;
    colToIdx[0] = 0;
    for (int i = 0; i < cols.size(); i++) {
      colToIdx[cols[i]] = i + 1;
    }
  }
  void add(int i, int x) {
    for (; i <= s; i += i & (-i)) bit[i] = (bit[i] + x);
  }
  ll sum(int i) {
    if (i == 0) return 0;
    ll r = 0;
    for (; i; i -= i & (-i)) {
      r = (r + bit[i]);
    }
    return r;
  }
  int idxForQuery(int i) {
    auto it = upper_bound(begin(cols), end(cols), i);
    if (it == cols.begin()) return 0;
    --it;
    return colToIdx[*it];
  }
  void insertValue(int col, int x) {
    int i = colToIdx[col];
    add(i, x);
  }
  ll getValue(int col) {
    int i = idxForQuery(col);
    return sum(i);
  }
};
int main() {
  scanf("%d", &n);
  a.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  Fen inv, cnt;
  inv.init(a);
  cnt.init(a);
  ll ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    ans += cnt.getValue(a[i] - 1);
    int c = inv.getValue(a[i] - 1);
    inv.insertValue(a[i], 1);
    cnt.insertValue(a[i], c);
  }
  cout << ans << '\n';
}

```