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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<map>
#include<math.h>

#define rep(i,a,n) for(int (i)=(a);(i)<(n);(i)++)
#define rrep(i,a,n) for(int (i)=(a);(i)>=(n);(i)--)

#define INF 1000000000
#define MOD 1000000007

typedef  long long ll;

using namespace std;

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

double euclid_distance(double a,double b){
    return sqrt(a*a+b*b);
}

int gcd(long long int a,long long int b){
    long long int r;
    if(a < b){
        int tmp;
        tmp = a;
        a = b;
        b = tmp;
    }
    while(r = a % b){
        a = b;
        b = r;
    }
    return a;
}

void Integer_factorization(long long int b,long long int n){
    long long int a = 2;
    long long int count = 0;
    long long int ans = 1;
    long long int ndash = n;
    if(b == 1) cout << n << endl;
    else{
    while(ndash >= a*a){
        if(n % a  == 0){
            count++;
            n/=a;

        if(count % b == 0){
            ans *= a;
        }
        }else{a++;count = 0;}
    }

    cout<<  ans << endl;
    }
}

void ys(){
    cout << "Yes" << endl;
}

void yb(){
    cout << "YES" << endl;
}

void ns(){
    cout << "No" << endl;
}

void nb(){
    cout << "NO" << endl;
}

void solve(void){
    int n,x;
    cin >> n >> x;
    int l[n];
    rep(i,0,n) cin >> l[i];
    int ans = 1;
    int d = l[0];
    int prev = 0;

    rep(i,1,n+1){
        
        if(d+prev <= x){
            ans++;
            prev = d + prev;
            if(i != n) d = l[i];
        }else{
            break;
        }
    }
    cout << ans << endl;

}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
}