Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows' heads.

Each cow i has a specified height hi (1 ≤ hi ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.

Consider this example:

        =
=       =
=   -   =         Cows facing right -->
=   =   =
= - = = =
= = = = = =
1 2 3 4 5 6

Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow's hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow's hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!

Let ci denote the number of cows whose hairstyle is visible from cow i; plea***pute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

Input

Line 1: The number of cows,  N. 
Lines 2..N+1: Line  i+1 contains a single integer that is the height of cow  i.

Output

Line 1: A single integer that is the sum of  c  1 through  cN.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

题意：
给定N个面向右边的奶牛，我们说一只奶牛a[i]，能看到另外一个牛奶的头发当且仅当 有一只奶牛a[j]瞒住，a[j]<a[i],并且j>i

思路：
裸的单调栈，维护一个从栈顶到栈底，严格不减小的栈。

细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rt return
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll n;
ll a[maxn];
ll r[maxn];
int main()
{
    gbtb;
    cin>>n;    
    repd(i,1,n)
    {
        cin>>a[i];
    }
    stack<int> s;
    for(int i=n;i>=1;i--)
    {
        while(s.size()&&a[s.top()]<a[i])
        {
            s.pop();
        }
        if(s.size())
        {
            r[i]=s.top();
        }else
        {
            r[i]=n+1;
        }
        s.push(i);
    }

    // repd(i,1,n)
    // {
    //     cout<<r[i]<<" ";
    // }
    // cout<<endl;

    ll ans=0ll;
    repd(i,1,n)
    {
        ans+=1ll*(r[i]-i-1);
        // db(r[i]-i-1);
    }
    cout<<ans<<endl;
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}