Problem:
Score : 200 points

Problem Statement

A ball will bounce along a number line, making N + 1 bounces. It will make the first bounce at coordinate D_1 = 0, and the i-th bounce (2 \leq i \leq N+1) at coordinate D_i = D_{i-1} + L_{i-1}.

How many times will the ball make a bounce where the coordinate is at most X?

Constraints

1 \leq N \leq 100

1 \leq L_i \leq 100

1 \leq X \leq 10000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X
L_1 L_2 ... L_{N-1} L_N

Output

Print the number of times the ball will make a bounce where the coordinate is at most X.

Sample Input 1

3 6
3 4 5

Sample Output 1

2

The ball will make a bounce at the coordinates 0, 3, 7 and 12, among which two are less than or equal to 6.

Sample Input 2

4 9
3 3 3 3

Sample Output 2

4

The ball will make a bounce at the coordinates 0, 3, 6, 9 and 12, among which four are less than or equal to 9.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

/*
String[]の初期化、整数の型がlongになっているか
関数（splitとかcontainsとか）内では"+"とかは正規表現となされるので"\\+"とする
配列のコピーはarr1=arr.clone()
HashSet<>[]はおのおの初期化した？？？？？
負の数の割り算は絶対値が小さいほうに切り捨てられるex.-1/2=0ので、ある値>kみたいな式だとバグらせやすいので注意
ある値>=kとして、切り上げた値をとるべき
toLowerCase()はアルファベット以外に対して行ってもエラーにはならない
ArrayDequeではpopとpushはStackと同じ操作になる
//import static java.lang.System.*; はしばらく使わなそう(使うとしてもexit(0)のためだけ)←return;でよさそう
最後にout.flush()またはout.close()をするのを忘れずに！
Point2D p2d=new Point2D.Double(d1,d2);
*/
//入力終了→Ctrl+D
import java.util.*;
import java.io.*;
import java.awt.*;
import java.awt.geom.Point2D;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        int n=sc.nextInt();
        int x=sc.nextInt();
        int[] l=new int[n];
        for (int i = 0; i < n; i++) {
            l[i]=sc.nextInt();
        }
        int[] d=new int[n+1];
        int ans=1;
        for (int i = 1; i < n + 1; i++) {
            d[i]=d[i-1]+l[i-1];
            if(d[i]<=x)ans++;
        }
        out.println(ans);
        out.close();
    }

    static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    //Union Find木(0-indexed)
    static class UF {
        static int[] d;

        static void Init(int size) {
            d = new int[size];
            for (int i = 0; i < size; i++) {
                d[i] = -1;
            }
        }

        static boolean unite(int x, int y) {
            x = find(x);
            y = find(y);
            if (x == y) return false;
            if (d[x] > d[y]) {
                int t = d[y];
                d[y] = d[x];
                d[x] = t;
            }
            d[x] += d[y];
            d[y] = x;
            return true;
        }

        static boolean same(int x, int y) {
            return find(x) == find(y);
        }

        static int find(int k) {
            return d[k] < 0 ? k : (d[k] = find(d[k]));
        }

        static int size(int k) {
            return -d[find(k)];
        }
    }

    static PrintWriter out = new PrintWriter(System.out);

    static class sc {
        static Scanner s = new Scanner(System.in);

        static String next() {
            return s.next();
        }

        static String nextLine() {
            return s.nextLine();
        }

        static int nextInt() {
            return Integer.parseInt(s.next());
        }

        static long nextLong() {
            return Long.parseLong(s.next());
        }

        static double nextDouble() {
            return Double.parseDouble(s.next());
        }

        static boolean hasNext() {
            return s.hasNext();
        }

        static boolean hasNextInt() {
            return s.hasNextInt();
        }

    }

}