Books

Description

When Valera has got some free time, he goes to the library to read some books. Today he's got t free minutes to read. That's why Valera took n books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to n. Valera needs a_i minutes to read the i-th book.

Valera decided to choose an arbitrary book with number i and read the books one by one, starting from this book. In other words, he will first read book number i, then book number i + 1, then book number i + 2 and so on. He continues the process until he either runs out of the free time or finishes reading the n-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.

Print the maximum number of books Valera can read.

Input

The first line contains two integers n and t(1 ≤ n ≤ 10⁵; 1 ≤ t ≤ 10⁹) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of n integers a₁, a₂, ..., a_n(1 ≤ a_i ≤ 10⁴), where number a_i shows the number of minutes that the boy needs to read the i-th book.

Output

Print a single integer — the maximum number of books Valera can read.

Sample Input

Input

4 5
3 1 2 1

Output

3

Input

3 3
2 2 3

Output

1


题意：给读n本书所花时间，book[i]表示读第i本书花的时间，问在不超过t的时间内最多可读多少本书。

思路：一道不需要算法的题，但还是有些技巧的，当时直接暴力结果超时了。
　　　设一个指针指向最长区间的起点，初始为0，每次尝试着加上book[i],若时间超过t,指针指向下一个起点。

#include<stdio.h>
#include<string.h>
#include<algorithm>
int book[100010];
int main()
{
    int n,t;
    while(~scanf("%d %d",&n,&t))
    {
        for(int i = 0; i < n; i++)
            scanf("%d",&book[i]);

        int books = 0;//最长区间。
        int start = 0;//指针指向初始化为第一个
        int sum = 0;
        int ans = -1;
        for(int i = 0; i < n; i++)
        {
            if(sum + book[i] <= t)
            {
                sum += book[i];
                books++;
                ans = std::max(ans,books);
            }
            else
            {
                i--;
                sum -= book[start];
                books--;
                start++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}