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)

/*
ID: thanhvq1
LANG: C++14
PROB: namenum
*/

#pragma GCC optimize ("O2")
#include<bits/stdc++.h>
#include<unistd.h>
using namespace std;
#define int long long
#define double long double
#define vec vector
#define vi vector<int>
#define FZ(n) memset((n),0,sizeof(n))
#define FMO(n) memset((n),-1,sizeof(n))
#define unset unordered_set
#define ii pair<int, int>
#define vii vector<ii>
#define F first
#define S second
#define PB push_back
#define ALL(x) begin(x),end(x)
#define SZ(x) ((int)(x).size())
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define FORE(i, a, b) for(int i = (a); i <= (b); i++)
#define FORD(i, a, b) for(int i = (a); i >= (b); i--)
#define bitcount(n) __builtin_popcountll(n)
template <typename T,typename U>
std::pair<T,U> operator+(const std::pair<T,U> & l,const std::pair<T,U> & r) {
  return {l.first+r.first, l.second+r.second};
}
typedef void (*callback_function)(void); // type for conciseness
const long long ZERO = 0LL;
const long long INF64 = 1e18;
const int INF32 = 1e9;
const int MOD = 1e9 + 7;
const double PI = acos(-1.0L);
const double EPS = static_cast<double>(1e-9);
typedef long long ll;

void FILEIO() {  freopen("in.txt", "r", stdin);  freopen("out.txt", "w", stdout); }
//void FILEIO() {  freopen("namenum.in", "r", stdin);  freopen("namenum.out", "w", stdout); }

int n, x;
vi l;

void solve() {
	cin >> n >> x;
	l.resize(n+1); FORE(i, 1, n) cin >> l[i];
	vi d(n+1); d[0]=0;
	int ans=1;
	FORE(i, 1, n) { d[i] = d[i-1] + l[i];
		if(d[i]<=x) ans++;
		else break;
	}
	cout << ans << endl;
}

//#undef int
signed main() {
//#define int long long
  ios_base::sync_with_stdio(false); cin.tie();
//  FILEIO();
//auto beginProgram = chrono::steady_clock::now();

// int t; cin >> t;
//  while (t--)
  {
		solve();
  }

  //auto endProgram = chrono::steady_clock::now();

  return 0;
}