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 tripletsi, 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

33 2 1

Output

1

Input

32 3 1

Output

0

Input

410 8 3 1

Output

4

Input

41 5 4 3

Output

1

题解：求出左端比该位置大的元素和右端比该位置小的元素。

#include<iostream>#include<stdio.h>#include<algorithm>#include<vector>typedef long long ll;const int Maxn=1e6+10;using namespace std;int n,sums[Maxn],suml[Maxn];ll l[Maxn],s[Maxn];struct nod{    int val,index;}a[Maxn];bool cmp1(nod a,nod b){    return a.val>b.val;}bool cmp2(nod a,nod b){    return a.val<b.val;}int lowbit(int x){    return x&(-x);}void updata(int x){    while(x<=n){        suml[x]++;        x+=lowbit(x);    }}void downdata(int x){    while(x>0){        sums[x]++;        x-=lowbit(x);    }}ll getSuml(int x){    ll ret=0;    while(x>0){        ret+=suml[x];        x-=lowbit(x);    }    return ret;}ll getSums(int x){    ll ret=0;    while(x<=n){        ret+=sums[x];        x+=lowbit(x);    }    return ret;}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d",&n)!=EOF){        for(int i=1;i<=n;i++) suml[i]=0,sums[i]=0;        for(int i=1;i<=n;i++) scanf("%d",&a[i].val),a[i].index=i;        sort(a+1,a+1+n,cmp1);        ll ans=0;        for(int i=1;i<=n;i++) l[a[i].index]=getSuml(a[i].index),updata(a[i].index);        sort(a+1,a+1+n,cmp2);        for(int i=1;i<=n;i++) s[a[i].index]=getSums(a[i].index),downdata(a[i].index);        for(int i=1;i<=n;i++) ans+=l[a[i].index]*s[a[i].index];        printf("%I64d\n",ans);    }}

这种方法简单一些。。。

#include<iostream>#include<stdio.h>#include<algorithm>#include<vector>typedef long long ll;const int Maxn=1e6+10;using namespace std;int n;struct nod{    int val,index;}a[Maxn];bool cmp(nod a,nod b){    return a.val>b.val;}int lowbit(int x){    return x&(-x);}struct st{    ll sum[Maxn];    void updata(int x,ll val){        while(x<=n){            sum[x]+=val;            x+=lowbit(x);        }    }    ll getSum(int x){        ll ret=0;        while(x>0){            ret+=sum[x];            x-=lowbit(x);        }        return ret;    }};st sum1,sum2;int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d",&n)!=EOF){        for(int i=1;i<=n;i++) sum1.sum[i]=0,sum2.sum[i]=0;        for(int i=1;i<=n;i++) scanf("%d",&a[i].val),a[i].index=i;        sort(a+1,a+1+n,cmp);        ll ans=0;        for(int i=1;i<=n;i++){            ll cnt1=sum1.getSum(a[i].index);//比该位置元素大个数            ll cnt2=sum2.getSum(a[i].index);//比该位置元素大对数            ans+=cnt2;            sum1.updata(a[i].index,1);            sum2.updata(a[i].index,cnt1);        }        printf("%I64d\n",ans);    }}