#include <bits/stdc++.h>
using namespace std;
long long a[1000005], b[1000005], sum[1000005 << 2], ls[1000005], rs[1000005];
void build(int rt, int l, int r) {
  sum[rt] = 0;
  if (l == r) {
    return;
  }
  build((rt << 1), l, ((r + l) >> 1));
  build((rt << 1 | 1), ((r + l) >> 1) + 1, r);
}
void updata(int rt, int l, int r, int x) {
  if (l == r) {
    sum[rt]++;
    return;
  }
  if (x <= b[((r + l) >> 1)])
    updata((rt << 1), l, ((r + l) >> 1), x);
  else
    updata((rt << 1 | 1), ((r + l) >> 1) + 1, r, x);
  sum[rt] = sum[(rt << 1)] + sum[(rt << 1 | 1)];
}
int query(int rt, int l, int r, int x) {
  if (b[r] < x) {
    return sum[rt];
  }
  if (b[l] >= x) return 0;
  int ans = 0;
  if (b[((r + l) >> 1)] < x)
    ans += (sum[(rt << 1)] + query((rt << 1 | 1), ((r + l) >> 1) + 1, r, x));
  else
    ans += query((rt << 1), l, ((r + l) >> 1), x);
  return ans;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%I64d", &a[i]);
      b[i] = a[i];
    }
    sort(b, b + n);
    build(1, 0, n - 1);
    for (int i = 0; i < n; i++) {
      ls[i] = i - query(1, 0, n - 1, a[i]);
      updata(1, 0, n - 1, a[i]);
    }
    build(1, 0, n - 1);
    for (int i = n - 1; i >= 0; i--) {
      rs[i] = query(1, 0, n - 1, a[i]);
      updata(1, 0, n - 1, a[i]);
    }
    long long ans = 0;
    for (int i = 1; i < n - 1; i++) {
      ans += (ls[i] * rs[i]);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
long long a[1000005], b[1000005], sum[1000005 << 2], ls[1000005], rs[1000005];
void build(int rt, int l, int r) {
  sum[rt] = 0;
  if (l == r) {
    return;
  }
  build((rt << 1), l, ((r + l) >> 1));
  build((rt << 1 | 1), ((r + l) >> 1) + 1, r);
}
void updata(int rt, int l, int r, int x) {
  if (l == r) {
    sum[rt]++;
    return;
  }
  if (x <= b[((r + l) >> 1)])
    updata((rt << 1), l, ((r + l) >> 1), x);
  else
    updata((rt << 1 | 1), ((r + l) >> 1) + 1, r, x);
  sum[rt] = sum[(rt << 1)] + sum[(rt << 1 | 1)];
}
int query(int rt, int l, int r, int x) {
  if (b[r] < x) {
    return sum[rt];
  }
  if (b[l] >= x) return 0;
  int ans = 0;
  if (b[((r + l) >> 1)] < x)
    ans += (sum[(rt << 1)] + query((rt << 1 | 1), ((r + l) >> 1) + 1, r, x));
  else
    ans += query((rt << 1), l, ((r + l) >> 1), x);
  return ans;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%I64d", &a[i]);
      b[i] = a[i];
    }
    sort(b, b + n);
    build(1, 0, n - 1);
    for (int i = 0; i < n; i++) {
      ls[i] = i - query(1, 0, n - 1, a[i]);
      updata(1, 0, n - 1, a[i]);
    }
    build(1, 0, n - 1);
    for (int i = n - 1; i >= 0; i--) {
      rs[i] = query(1, 0, n - 1, a[i]);
      updata(1, 0, n - 1, a[i]);
    }
    long long ans = 0;
    for (int i = 1; i < n - 1; i++) {
      ans += (ls[i] * rs[i]);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```