CodeForces 61E Enemy is weak

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 a_i > a_j > a_k where a_x 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 a_i (1 ≤ i ≤ n, 1 ≤ a_i ≤ 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).

Example

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

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int n;
int sum[1000050];
int a[1000050];
struct node{
    int num;
    int power;
}p[1000050];
bool cmp(node a,node b)
{
    return a.power<b.power;
}
bool cmp2(node a,node b)
{
    return a.num<b.num;
}
int lowbite(int x)
{
    return x&(-x);
}
int update(int x)
{
    while(x<=n)
    {
        sum[x]++;
        x+=lowbite(x);
    }
}
long long int find(int x,int i)
{
    long long int  ans=0;
    for(int t=x;t>0;t-=lowbite(t))
    {
        ans+=sum[t];
    }
    long long int  qnx=i-ans;
    return  qnx*(x-ans);

}
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&p[i].power);
        p[i].num=i;
    }
    sort(p+1,p+1+n,cmp);
    for(int i=1;i<=n;i++)
    {
        p[i].power=i;
    }
    sort(p+1,p+n+1,cmp2);
    long long int ans=0;
    for(int i=1;i<n;i++)
    {
        update(p[i].power);
        ans+=find(p[i].power,i);
    }
    cout<<ans<<endl;
    return 0;
}