Codeforces B. Cards Sorting 【瞎搞】

B. Cards Sorting

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this integer is between 1 and 100 000, inclusive. It is possible that some cards have the same integers on them.

Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn't know where this card (or these cards) is.

You are to determine the total number of times Vasily takes the top card from the deck.

Input

The first line contains single integer n (1 ≤ n ≤ 100 000) — the number of cards in the deck.

The second line contains a sequence of n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000), where ai is the number written on the i-th from top card in the deck.

Output

Print the total number of times Vasily takes the top card from the deck.

Examples

input

46 3 1 2

output

7

input

11000

output

1

input

73 3 3 3 3 3 3

output

7

Note

In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2, 6, 3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6, 3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards.

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假设现在剩下n张牌,取出n张牌进行操作  等价于  丢掉牌堆里最小的牌x,如果存在第二小的y,而且y在最后一张x后面,则把最后一张x后面的y也丢掉

例如 2 2 1 2 2 3

n次操作后剩下 2 2 3

如此这般...模拟一遍就是了

#include<stdio.h>#include<bits/stdc++.h>#define ll long long#define pii pair<int,int>#define MEM(a,x) memset(a,x,sizeof(a))#define lowbit(x) ((x)&-(x))using namespace std;const int inf=0x3f3f3f3f;const int N = 1e6 + 5;int a[N];//valvector<int>pos[N];int maxPos[N];ll slove(int n){    sort(a,a+n);    int tn=n;//牌堆剩余牌数    ll ans=tn;//需要的操作次数    tn-=pos[a[0]].size();//扔掉所有a[0]    pos[a[0]].clear();    for(int i=1;i<n;++i){        if(pos[a[i]].empty()){            continue;        }        if(pos[a[i]].back()>maxPos[a[i-1]]){//删去最后一个a[i-1]后面的a[i]            vector<int>&vec=pos[a[i]];            auto it=upper_bound(vec.begin(),vec.end(),maxPos[a[i-1]]);            int num=vec.end()-it;            vec.erase(it,vec.end());            if(!vec.empty()){                maxPos[a[i]]=vec.back();            }            tn-=num;        }        else{//取出整个牌堆的牌,丢掉a[i]            ans+=tn;            tn-=pos[a[i]].size();            pos[a[i]].clear();        }    }    return ans;}int main(){    //freopen("/home/lu/code/r.txt","r",stdin);    int n;    while(~scanf("%d",&n)){        for(int i=0;i<N;++i){            pos[i].clear();        }        for(int i=0;i<n;++i){            scanf("%d",&a[i]);            pos[a[i]].push_back(i);            maxPos[a[i]]=i;        }        printf("%lld\n",slove(n));    }    return 0;}