Everybody in Russia uses Gregorian calendar. In this calendar there are 31 days in January, 28 or 29 days in February (depending on whether the year is leap or not), 31 days in March, 30 days in April, 31 days in May, 30 in June, 31 in July, 31 in August, 30 in September, 31 in October, 30 in November, 31 in December.
A year is leap in one of two cases: either its number is divisible by 4, but not divisible by 100, or is divisible by 400. For example, the following years are leap: 2000, 2004, but years 1900 and 2018 are not leap.
In this problem you are given n (1 ≤ n ≤ 24) integers a1, a2, ..., an, and you have to check if these integers could be durations in days of n consecutive months, according to Gregorian calendar. Note that these months could belong to several consecutive years. In other words, check if there is a month in some year, such that its duration is a1 days, duration of the next month is a2 days, and so on.
Input
The first line contains single integer n (1 ≤ n ≤ 24) — the number of integers.
The second line contains n integers a1, a2, ..., an (28 ≤ ai ≤ 31) — the numbers you are to check.
Output
If there are several consecutive months that fit the sequence, print "YES" (without quotes). Otherwise, print "NO" (without quotes).
You can print each letter in arbitrary case (small or large).
Examples
Input
4 31 31 30 31
Output
Yes
Input
2 30 30
Output
No
Input
5 29 31 30 31 30
Output
Yes
Input
3 31 28 30
Output
No
Input
3 31 31 28
Output
Yes
Note
In the first example the integers can denote months July, August, September and October.
In the second example the answer is no, because there are no two consecutive months each having 30 days.
In the third example the months are: February (leap year) — March — April – May — June.
In the fourth example the number of days in the second month is 28, so this is February. March follows February and has 31 days, but not 30, so the answer is NO.
In the fifth example the months are: December — January — February (non-leap year).
呃。比赛时做的一道水题，但是wa了好几发，总是卡在第14组样例，当时就想着直接暴力，完全没毛病啊，后来发现虽然题目给的n是小于等于24的。但是不一定是恰巧从第年的一月到第二年的12月（因为当时我mon数组里存的是24个月份），后来mon数组改成36个月，果然a了，其实mon数组可以开一个二维的，但是当时不知道怎么没开对，后来也懒得改了。。
还要注意的一点是现实中是不可能存在两个连续的润年的。
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
typedef long long LL;
int mon1[500]={31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[500]={31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31};
int mon3[500]={31,29,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31};
int mon4[500]={31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,31,30,31};
queue<int> q1,q2,q3;
int main()
{
int n,i,j,k,g=0;
int a[100];
cin>>n;
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<36;i++)
{
int e=0;
for(j=i;j<n+i;j++)
{
if(mon1[j]!=a[e++])
break;
}
if(j>=n+i)
g++;
}
for(i=0;i<36;i++)
{
int e=0;
for(j=i;j<n+i;j++)
{
if(mon2[j]!=a[e++])
break;
}
if(j>=n+i)
g++;
}
for(i=0;i<36;i++)
{
int e=0;
for(j=i;j<n+i;j++)
{
if(mon3[j]!=a[e++])
break;
}
if(j>=n+i)
g++;
}
for(i=0;i<36;i++)
{
int e=0;
for(j=i;j<n+i;j++)
{
if(mon4[j]!=a[e++])
break;
}
if(j>=n+i)
g++;
}
if(g!=0)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
return 0;
}
