#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
vector<pair<int, int> > V, Q;
long long acc[N], H[N], L[N];
void update(int id) {
  for (int i = id; i < N; i = i + (i & -i)) {
    acc[i]++;
  }
}
void build(int sz) {
  for (int i = 1; i <= sz; i++) {
    update(i);
  }
}
long long query(int id) {
  long long sum = 0;
  for (int i = id; i > 0; i = i - (i & -i)) {
    sum = sum + acc[i];
  }
  return sum;
}
int main() {
  int n, x;
  while (scanf("%d", &n) != EOF) {
    V.clear();
    for (int i = 0; i < n; i++) {
      scanf("%d", &x);
      V.push_back(pair<int, int>(x, i + 1));
    }
    sort(V.begin(), V.end());
    memset(acc, 0, sizeof acc);
    for (int i = 0; i < V.size(); i++) {
      Q.clear();
      Q.push_back(V[i]);
      i++;
      while (i < V.size() and Q[0].first == V[i].first) {
        Q.push_back(V[i]);
        i++;
      }
      i--;
      for (int j = 0; j < Q.size(); j++) {
        H[Q[j].second] = query(n) - query(Q[j].second);
      }
      for (int j = 0; j < Q.size(); j++) {
        update(Q[j].second);
      }
    }
    memset(acc, 0, sizeof acc);
    for (int i = V.size() - 1; i >= 0; i--) {
      Q.clear();
      Q.push_back(V[i]);
      i--;
      while (i >= 0 and Q[0].first == V[i].first) {
        Q.push_back(V[i]);
        i--;
      }
      i++;
      for (int j = 0; j < Q.size(); j++) {
        L[Q[j].second] = query(Q[j].second);
      }
      for (int j = 0; j < Q.size(); j++) {
        update(Q[j].second);
      }
    }
    long long ans = 0LL;
    for (int i = 1; i <= n; i++) {
      ans = ans + H[i] * L[i];
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 1e6 + 10;
vector<pair<int, int> > V, Q;
long long acc[N], H[N], L[N];
void update(int id) {
  for (int i = id; i < N; i = i + (i & -i)) {
    acc[i]++;
  }
}
void build(int sz) {
  for (int i = 1; i <= sz; i++) {
    update(i);
  }
}
long long query(int id) {
  long long sum = 0;
  for (int i = id; i > 0; i = i - (i & -i)) {
    sum = sum + acc[i];
  }
  return sum;
}
int main() {
  int n, x;
  while (scanf("%d", &n) != EOF) {
    V.clear();
    for (int i = 0; i < n; i++) {
      scanf("%d", &x);
      V.push_back(pair<int, int>(x, i + 1));
    }
    sort(V.begin(), V.end());
    memset(acc, 0, sizeof acc);
    for (int i = 0; i < V.size(); i++) {
      Q.clear();
      Q.push_back(V[i]);
      i++;
      while (i < V.size() and Q[0].first == V[i].first) {
        Q.push_back(V[i]);
        i++;
      }
      i--;
      for (int j = 0; j < Q.size(); j++) {
        H[Q[j].second] = query(n) - query(Q[j].second);
      }
      for (int j = 0; j < Q.size(); j++) {
        update(Q[j].second);
      }
    }
    memset(acc, 0, sizeof acc);
    for (int i = V.size() - 1; i >= 0; i--) {
      Q.clear();
      Q.push_back(V[i]);
      i--;
      while (i >= 0 and Q[0].first == V[i].first) {
        Q.push_back(V[i]);
        i--;
      }
      i++;
      for (int j = 0; j < Q.size(); j++) {
        L[Q[j].second] = query(Q[j].second);
      }
      for (int j = 0; j < Q.size(); j++) {
        update(Q[j].second);
      }
    }
    long long ans = 0LL;
    for (int i = 1; i <= n; i++) {
      ans = ans + H[i] * L[i];
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```