#include <bits/stdc++.h>
int a[1000000], p[1000000], b[1000000], c[1000000], c2[1000000];
void count(int l, int r) {
  if (l < r) {
    int m = (l + r) / 2;
    count(l, m);
    count(m + 1, r);
    int i = l, j = m + 1, k = l;
    while (i <= m && j <= r)
      if (a[p[i]] > a[p[j]]) {
        c[p[i]] += r - j + 1;
        b[k++] = p[i++];
      } else {
        b[k++] = p[j++];
      }
    while (i <= m) b[k++] = p[i++];
    while (j <= r) b[k++] = p[j++];
    for (k = l; k <= r; k++) p[k] = b[k];
  }
}
void count2(int l, int r) {
  if (l < r) {
    int m = (l + r) / 2;
    count2(l, m);
    count2(m + 1, r);
    int i = l, j = m + 1, k = l;
    while (i <= m && j <= r)
      if (a[p[j]] < a[p[i]]) {
        c2[p[j]] += m - i + 1;
        b[k++] = p[j++];
      } else {
        b[k++] = p[i++];
      }
    while (i <= m) b[k++] = p[i++];
    while (j <= r) b[k++] = p[j++];
    for (k = l; k <= r; k++) p[k] = b[k];
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int i;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    p[i] = i;
  }
  memset(c, 0, sizeof(c));
  count(0, n - 1);
  memset(c2, 0, sizeof(c2));
  for (i = 0; i < n; i++) p[i] = i;
  count2(0, n - 1);
  long long t = 0;
  for (i = 0; i < n; i++) t += 1LL * c[i] * c2[i];
  printf("%I64d\n", t);
  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>
int a[1000000], p[1000000], b[1000000], c[1000000], c2[1000000];
void count(int l, int r) {
  if (l < r) {
    int m = (l + r) / 2;
    count(l, m);
    count(m + 1, r);
    int i = l, j = m + 1, k = l;
    while (i <= m && j <= r)
      if (a[p[i]] > a[p[j]]) {
        c[p[i]] += r - j + 1;
        b[k++] = p[i++];
      } else {
        b[k++] = p[j++];
      }
    while (i <= m) b[k++] = p[i++];
    while (j <= r) b[k++] = p[j++];
    for (k = l; k <= r; k++) p[k] = b[k];
  }
}
void count2(int l, int r) {
  if (l < r) {
    int m = (l + r) / 2;
    count2(l, m);
    count2(m + 1, r);
    int i = l, j = m + 1, k = l;
    while (i <= m && j <= r)
      if (a[p[j]] < a[p[i]]) {
        c2[p[j]] += m - i + 1;
        b[k++] = p[j++];
      } else {
        b[k++] = p[i++];
      }
    while (i <= m) b[k++] = p[i++];
    while (j <= r) b[k++] = p[j++];
    for (k = l; k <= r; k++) p[k] = b[k];
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int i;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    p[i] = i;
  }
  memset(c, 0, sizeof(c));
  count(0, n - 1);
  memset(c2, 0, sizeof(c2));
  for (i = 0; i < n; i++) p[i] = i;
  count2(0, n - 1);
  long long t = 0;
  for (i = 0; i < n; i++) t += 1LL * c[i] * c2[i];
  printf("%I64d\n", t);
  return 0;
}

```