題干：

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

題目大意：

? ?就是說n個人排隊來買票，票價25元，每個人都出 25,50,100中的一種鈔票，售票處需要找零錢，，問能不能找錢成功。。

?

解題報告：

? ? 分情況亂搞就可以了。

AC代碼：

#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 ;}

?

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