题干：

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

4
25 25 50 50

Output

YES

Input

2
25 100

Output

NO

Input

4
50 50 25 25

Output

NO

题目大意：

   就是说n个人排队来买票，票价25元，每个人都出 25,50,100中的一种钞票，售票处需要找零钱，，问能不能找钱成功。。

 

解题报告：

    分情况乱搞就可以了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int bk[500];
int main()
{
	int n,tmp,flag = 1;
	cin>>n;
	for(int i = 1; i<=n; i++) {
		scanf("%d",&tmp);
		bk[tmp]++;
		if(flag == 0) continue;
		if(tmp == 25) continue;
		if(tmp == 50) {
			if(bk[25] != 0) {
				bk[25]--;continue;
			}
		}
		if(tmp == 100) {
			if(bk[50]>=1 && bk[25]>=1) {
				bk[50]--;bk[25]--;
				continue;
			}
			if(bk[25]>=3) {
				bk[25]-=3;
				continue;
			}
		}
		flag = 0;
	}
	if(flag) puts("YES");
	else puts("NO");

	return 0 ;
 }