A. Cinema Line-优快云博客

本文介绍了一种解决电影院售票问题的算法，通过处理不同面额的纸币（25、50、100），确保每位顾客都能顺利购买到票价为25的电影票，并在没有初始资金的情况下进行找零。算法考虑了顾客排队购票的顺序，通过跟踪25元和50元的票据数量，判断是否能完成所有交易。

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The new "Die Hard" movie has just been released! There are n people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of people in the line. The next line contains n integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.

Output

Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".

Examples

input

Copy

4
25 25 50 50

output

Copy

YES

input

Copy

2
25 100

output

Copy

NO

input

Copy

4
50 50 25 25

output

Copy

NO

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
    int n,a[100005];
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    int t1=0,t2=0,flag=1;
    for(int i=0;i<n;i++)
    {
        if(a[i]==25)
        {
            t1++;
            continue;
        }
        if(a[i]==50)
        {
            if(t1>=1)
            {
                t2++;
                t1--;
                continue;
            }
            else
            {
                flag=0;
                break;
            }
        }
        if(a[i]==100)
        {
            if(t1>=1&&t2>=1)
            {
                t1--;
                t2--;
                continue;
            }
            if(t1>=3)
            {
                t1=t1-3;
                continue;
            }

            else
            {
                flag=0;
                break;
            }
        }

    }
    if(flag==1)
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
    return 0;
}