#include <bits/stdc++.h>
using namespace std;
long long val[1000010 << 2][3];
int a[1000010];
int v[1000010];
int find(int x, int n) {
  int left = 1, right = n - 1;
  while (left != right + 1) {
    int mid = (left + right) >> 1;
    if (v[mid] == x) {
      return mid;
    }
    if (v[mid] < x) {
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return -1;
}
void update(int left, int right, int id, int op, long long v, int idx) {
  if (left == right) {
    val[idx][op] += v;
    return;
  }
  int mid = (left + right) >> 1;
  if (mid < id) {
    update(mid + 1, right, id, op, v, idx << 1 | 1);
  } else {
    update(left, mid, id, op, v, idx << 1);
  }
  val[idx][op] = val[idx << 1][op] + val[idx << 1 | 1][op];
}
long long query(int left, int right, int L, int R, int op, int idx) {
  if (left >= L && right <= R) {
    return val[idx][op];
  }
  int mid = (left + right) >> 1;
  if (L > mid) {
    return query(mid + 1, right, L, R, op, idx << 1 | 1);
  } else if (mid >= R) {
    return query(left, mid, L, R, op, idx << 1);
  } else {
    return query(left, mid, L, mid, op, idx << 1) +
           query(mid + 1, right, mid + 1, R, op, idx << 1 | 1);
  }
}
int main(void) {
  int n, k = 1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    v[k++] = a[i];
  }
  sort(v + 1, v + k);
  int m = 2;
  for (int i = 2; i <= n; i++) {
    if (v[i] != v[i - 1]) {
      v[m++] = v[i];
    }
  }
  update(1, m, find(a[1], m), 0, 1, 1);
  long long ans = 0;
  for (int i = 2; i <= n; i++) {
    update(1, m, find(a[i], m), 0, 1, 1);
    for (int j = 0; j < 2; j++) {
      long long x = query(1, m, find(a[i], m) + 1, m, j, 1);
      if (j == 1) {
        ans += x;
      }
      update(1, m, find(a[i], m), j + 1, x, 1);
    }
  }
  cout << ans << endl;
  return 0;
}

### 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;
long long val[1000010 << 2][3];
int a[1000010];
int v[1000010];
int find(int x, int n) {
  int left = 1, right = n - 1;
  while (left != right + 1) {
    int mid = (left + right) >> 1;
    if (v[mid] == x) {
      return mid;
    }
    if (v[mid] < x) {
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return -1;
}
void update(int left, int right, int id, int op, long long v, int idx) {
  if (left == right) {
    val[idx][op] += v;
    return;
  }
  int mid = (left + right) >> 1;
  if (mid < id) {
    update(mid + 1, right, id, op, v, idx << 1 | 1);
  } else {
    update(left, mid, id, op, v, idx << 1);
  }
  val[idx][op] = val[idx << 1][op] + val[idx << 1 | 1][op];
}
long long query(int left, int right, int L, int R, int op, int idx) {
  if (left >= L && right <= R) {
    return val[idx][op];
  }
  int mid = (left + right) >> 1;
  if (L > mid) {
    return query(mid + 1, right, L, R, op, idx << 1 | 1);
  } else if (mid >= R) {
    return query(left, mid, L, R, op, idx << 1);
  } else {
    return query(left, mid, L, mid, op, idx << 1) +
           query(mid + 1, right, mid + 1, R, op, idx << 1 | 1);
  }
}
int main(void) {
  int n, k = 1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    v[k++] = a[i];
  }
  sort(v + 1, v + k);
  int m = 2;
  for (int i = 2; i <= n; i++) {
    if (v[i] != v[i - 1]) {
      v[m++] = v[i];
    }
  }
  update(1, m, find(a[1], m), 0, 1, 1);
  long long ans = 0;
  for (int i = 2; i <= n; i++) {
    update(1, m, find(a[i], m), 0, 1, 1);
    for (int j = 0; j < 2; j++) {
      long long x = query(1, m, find(a[i], m) + 1, m, j, 1);
      if (j == 1) {
        ans += x;
      }
      update(1, m, find(a[i], m), j + 1, x, 1);
    }
  }
  cout << ans << endl;
  return 0;
}

```