Educational Codeforces Round 31 A. Book Reading【暴力】

A. Book Reading

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently Luba bought a very interesting book. She knows that it will take t seconds to read the book. Luba wants to finish reading as fast as she can.

But she has some work to do in each of n next days. The number of seconds that Luba has to spend working during i-th day is ai. If some free time remains, she can spend it on reading.

Help Luba to determine the minimum number of day when she finishes reading.

It is guaranteed that the answer doesn't exceed n.

Remember that there are 86400 seconds in a day.

Input

The first line contains two integers n and t (1 ≤ n ≤ 100, 1 ≤ t ≤ 106) — the number of days and the time required to read the book.

The second line contains n integers ai (0 ≤ ai ≤ 86400) — the time Luba has to spend on her work during i-th day.

Output

Print the minimum day Luba can finish reading the book.

It is guaranteed that answer doesn't exceed n.

Examples

input

2 2
86400 86398

output

2

input

2 86400
0 86400

output

1
【题意】：给出总天数n，读书时间t。n天内的干活时间（按秒记，一天86400秒）。求最少的读完书的天数。
【分析】：暴力。
【代码】：

#include <bits/stdc++.h>

using namespace std;
#define inf 1e18+100
#define LL long long

const int maxn = 86400+100;

int main()
{
    LL n,t,res; //给定时间，读书时间
    LL r;
    LL a[maxn];//干活时间 求花几天读书
    while(cin>>n>>t)//2 86400 //1 86399
   {
       res=0;
       for(int i=0;i<n;i++)
       {
           cin>>a[i];
           res+=(86400-a[i]);//86400-干活时间=留下的读书时间
           if(res>=t)//剩余读书时间》需求读书时间，则可以于这天读完，直接输出该天是第几天
           {
               cout<<i+1<<endl;
               return 0;
           }
       }
   }
}