#include <bits/stdc++.h>
using namespace std;
const int maxn = 1111111;
int maxm[maxn], a[maxn], x[maxn];
long long sum[maxn << 2];
void pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
void build(int l, int r, int rt) {
  sum[rt] = 0;
  if (l == r) return;
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
}
void update(int p, int l, int r, int rt) {
  if (p == l && r == p) {
    sum[rt]++;
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    update(p, l, m, rt << 1);
  else
    update(p, m + 1, r, rt << 1 | 1);
  pushup(rt);
}
long long query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return sum[rt];
  int m = (l + r) >> 1;
  long long ret = 0;
  if (L <= m) ret += query(L, R, l, m, rt << 1);
  if (R > m) ret += query(L, R, m + 1, r, rt << 1 | 1);
  return ret;
}
int Bin(int key, int l, int r) {
  int mid = 0;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (x[mid] == key) return mid;
    if (x[mid] < key)
      l = mid + 1;
    else
      r = mid - 1;
  }
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    int nn = 1;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      x[i] = a[i];
    }
    sort(x + 1, x + n + 1);
    for (int i = 2; i <= n; ++i)
      if (x[i] != x[i - 1]) x[++nn] = x[i];
    build(1, nn, 1);
    for (int i = 1; i <= n; ++i) {
      int tmp = Bin(a[i], 1, nn);
      if (i > 1 && i < n) {
        maxm[i] = query(tmp, nn, 1, nn, 1);
      }
      update(tmp, 1, nn, 1);
    }
    build(1, nn, 1);
    long long ans = 0;
    for (int i = n; i > 1; --i) {
      int tmp = Bin(a[i], 1, nn);
      if (i != n) ans += query(1, tmp, 1, nn, 1) * maxm[i];
      update(tmp, 1, nn, 1);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Generate 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 maxn = 1111111;
int maxm[maxn], a[maxn], x[maxn];
long long sum[maxn << 2];
void pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
void build(int l, int r, int rt) {
  sum[rt] = 0;
  if (l == r) return;
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
}
void update(int p, int l, int r, int rt) {
  if (p == l && r == p) {
    sum[rt]++;
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    update(p, l, m, rt << 1);
  else
    update(p, m + 1, r, rt << 1 | 1);
  pushup(rt);
}
long long query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return sum[rt];
  int m = (l + r) >> 1;
  long long ret = 0;
  if (L <= m) ret += query(L, R, l, m, rt << 1);
  if (R > m) ret += query(L, R, m + 1, r, rt << 1 | 1);
  return ret;
}
int Bin(int key, int l, int r) {
  int mid = 0;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (x[mid] == key) return mid;
    if (x[mid] < key)
      l = mid + 1;
    else
      r = mid - 1;
  }
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    int nn = 1;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      x[i] = a[i];
    }
    sort(x + 1, x + n + 1);
    for (int i = 2; i <= n; ++i)
      if (x[i] != x[i - 1]) x[++nn] = x[i];
    build(1, nn, 1);
    for (int i = 1; i <= n; ++i) {
      int tmp = Bin(a[i], 1, nn);
      if (i > 1 && i < n) {
        maxm[i] = query(tmp, nn, 1, nn, 1);
      }
      update(tmp, 1, nn, 1);
    }
    build(1, nn, 1);
    long long ans = 0;
    for (int i = n; i > 1; --i) {
      int tmp = Bin(a[i], 1, nn);
      if (i != n) ans += query(1, tmp, 1, nn, 1) * maxm[i];
      update(tmp, 1, nn, 1);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```