Gym 100989H

After the data structures exam, students lined up in the cafeteria to have a drink and chat about how much they have enjoyed the exam and how good their professors are. Since it was late in the evening, the cashier has already closed the cash register and does not have any change with him.

The students are going to pay using Jordanian money notes, which are of the following types: 1, 5, 10, 20, 50.

Given how much each student has to pay, the set of notes he’s going to pay with, and the order in which the students arrive at the cashier, your task is to find out if the cashier will have enough change to return to each of the student when they arrive at the cashier.

Input

The first line of input contains a single integer N (1 ≤ N ≤ 10⁵), the number of students in the queue.

Each of the following N lines describes a student and contains 6 integers, K, F₁, F₂, F₃, F₄, and F₅, where K represents the amount of money the student has to pay, and F_i (0 ≤ F_i ≤ 100) represents the amount of the i^th type of money this student is going to give to the cashier.

The students are given in order; the first student is in front of the cashier.

It is guaranteed that no student will pay any extra notes. In other words, after removing any note from the set the student is going to give to the cashier, the amount of money will be less than what is required to buy the drink.

Output

Print yes if the cashier will have enough change to return to each of the students when they arrive in the given order, otherwise print no.

Examples

Input

3
4 0 1 0 0 0
9 4 1 0 0 0
8 0 0 1 0 0

Output

no

Input

3
9 4 1 0 0 0
4 0 1 0 0 0
8 0 0 1 0 0

Output

yes

题意：n个人买东西去柜台是否能全部找钱，k表示要付的钱，后面的数对应钞票1,5,10,20,50的数量，起初柜台里是没钱的。

思路：必须先判断能不能找零，之后才能拿钱，一开始没注意一直卡在第九组数据。如果第一个人给的钱不能刚好支付完，则没有零钱找给他，所以这种情况就直接no，后面假装输入一下就可以continue掉。如果需要找零钱，先找面值最大的，如果最后能找完，则收下他付的钱。
代码如下：

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
	int n;
	int s[5]={1,5,10,20,50};
	int v[5]={0};
	int a[5];
	long long int k;
	cin>>n;
	int flag=0,c=0;
	while(n--)
	{
		c++;
		cin>>k;
		long long int t=0;
		for(int i=0;i<5;i++)
		{
			cin>>a[i];
			t+=a[i]*s[i];
		}
		if(flag)
		continue;
		k=t-k;
		if(k==0)
		{
			for(int i=0;i<5;i++)
			v[i]+=a[i];
		}
		else
		{
			if(c==1)
			{
				flag=1;
				continue;
			}
			while(k>0)
			{
				while(k>=s[4]&&v[4])
				{
					k-=s[4];
					v[4]--;
				}
				if(k==0)
				break;
				while(k>=s[3]&&v[3])
				{
					k-=s[3];
					v[3]--;
				}
				if(k==0)
				break;
				while(k>=s[2]&&v[2])
				{
					k-=s[2];
					v[2]--;
				}
				if(k==0)
				break;
				while(k>=s[1]&&v[1])
				{
					k-=s[1];
					v[1]--;
				}
				if(k==0)
				break;
				while(k>=s[0]&&v[0])
				{
					k-=s[0];
					v[0]--;
				}
				if(k==0)
				break;
				if(k>0)
				{
					flag=1;
					break;
				}
			}
			if(flag==0)
			{
				for(int i=0;i<5;i++)
				v[i]+=a[i];
			}
		}
	}
	if(flag==1)
	cout<<"no"<<endl;
	else
	cout<<"yes"<<endl;
	return 0;
}