#include <bits/stdc++.h>
using namespace std;
class ST {
 private:
  vector<long long> st;
  int n;
  int left(int p) { return p << 1; }
  int right(int p) { return left(p) + 1; }
  int query(int p, int L, int R, int i) {
    if (L > i) return 0;
    if (R <= i) return st[p];
    return query(left(p), L, (L + R) >> 1, i) +
           query(right(p), ((L + R) >> 1) + 1, R, i);
  }
  int update(int p, int L, int R, int idx) {
    if (L > idx || R < idx) return st[p];
    if (L == R) {
      return ++st[p];
    }
    return st[p] = update(left(p), L, (L + R) >> 1, idx) +
                   update(right(p), ((L + R) >> 1) + 1, R, idx);
  }

 public:
  ST(int _n) {
    n = _n;
    st.assign(4 * n, 0);
  }
  int query(int i) {
    int ret = query(1, 0, n - 1, i);
    update(i);
    return ret;
  }
  void update(int idx) { update(1, 0, n - 1, idx); }
};
int main() {
  int n;
  map<long long, long long> mp;
  long long ans = 0;
  scanf("%I64d", &n);
  vector<long long> vec(n), vec2(n);
  for (long long i = 0; i < n; i++) scanf("%I64d", &vec[i]), vec2[i] = vec[i];
  ST st(n);
  sort(vec2.begin(), vec2.end());
  for (long long i = 0; i < n; i++) mp[vec2[i]] = i;
  for (long long i = 0; i < n - 1; i++) {
    long long q = st.query(mp[vec[i]]);
    ans += (i - q) * (mp[vec[i]] - q);
  }
  printf("%I64d", ans);
  return 0;
}

### Prompt

Please formulate 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;
class ST {
 private:
  vector<long long> st;
  int n;
  int left(int p) { return p << 1; }
  int right(int p) { return left(p) + 1; }
  int query(int p, int L, int R, int i) {
    if (L > i) return 0;
    if (R <= i) return st[p];
    return query(left(p), L, (L + R) >> 1, i) +
           query(right(p), ((L + R) >> 1) + 1, R, i);
  }
  int update(int p, int L, int R, int idx) {
    if (L > idx || R < idx) return st[p];
    if (L == R) {
      return ++st[p];
    }
    return st[p] = update(left(p), L, (L + R) >> 1, idx) +
                   update(right(p), ((L + R) >> 1) + 1, R, idx);
  }

 public:
  ST(int _n) {
    n = _n;
    st.assign(4 * n, 0);
  }
  int query(int i) {
    int ret = query(1, 0, n - 1, i);
    update(i);
    return ret;
  }
  void update(int idx) { update(1, 0, n - 1, idx); }
};
int main() {
  int n;
  map<long long, long long> mp;
  long long ans = 0;
  scanf("%I64d", &n);
  vector<long long> vec(n), vec2(n);
  for (long long i = 0; i < n; i++) scanf("%I64d", &vec[i]), vec2[i] = vec[i];
  ST st(n);
  sort(vec2.begin(), vec2.end());
  for (long long i = 0; i < n; i++) mp[vec2[i]] = i;
  for (long long i = 0; i < n - 1; i++) {
    long long q = st.query(mp[vec[i]]);
    ans += (i - q) * (mp[vec[i]] - q);
  }
  printf("%I64d", ans);
  return 0;
}

```