Good Bye 2016 A. New Year and Hurry【贪心/做题目每道题花费时间按步长为5等差增长，求剩余时间够做几道题】

A. New Year and Hurry

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be n problems, sorted by difficulty, i.e. problem 1 is the easiest and problem n is the hardest. Limak knows it will take him 5·i minutes to solve the i-th problem.

Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs k minutes to get there from his house, where he will participate in the contest first.

How many problems can Limak solve if he wants to make it to the party?

Input

The only line of the input contains two integers n and k (1 ≤ n ≤ 10, 1 ≤ k ≤ 240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.

Output

Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.

Examples

input

3 222

output

2

input

4 190

output

4

input

7 1

output

7

Note

In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3problems so the answer is 2.

In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.

In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7problems.

【代码】：

#include <bits/stdc++.h>

using namespace std;
int main()
{
    int n,k,ans,res,cnt;
    while(cin>>n>>k)
    {
        ans=0,cnt=0;
        res=240-k;
        //cout<<"res= "<<res<<endl;//50
        for(int i=5;;i+=5)
        {
            ans+=i;
            if(ans>res||cnt>=n)
            {
                break;
            }
            cnt++;
           // printf("cnt=%d ans=%d
",cnt,ans);//
        }
        cout<<cnt<<endl;
    }
    return 0;
}

#include<iostream>  
#include<cstdio>  
#include<cmath>  
using namespace std;  
int main()  
{  
    int n,k;  
    while(~scanf("%d%d",&n,&k))  
    {  
        int i,sum=k;  
        for(i=0;i<=n;i++)  
        {  
            sum+=i*5;  
            if(sum>240)  
                break;  
        }  
        cout<<i-1<<endl;  
    }  
    return 0;  
}