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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <random>
#include <iomanip>

using namespace std;
typedef long long unsigned int ll;
typedef pair<int, int> P;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))

#define fillInt(xs, x)        \
  for (int i = 0; i < (x); i++) \
    scanf("%d", &xs[i]);
#define fillLong(xs, x)       \
  for (int i = 0; i < (x); i++) \
    scanf("%ld", &xs[i]);
#define fillString(xs, x)       \
  for (int i = 0; i < (x); i++) \
    cin >> xs[i];
#define sortv(xs) sort(xs.begin(), xs.end())
#define sortvdi(xs) sort(xs.begin(), xs.end(), std::greater<int>())
#define lbv(xs, x) lower_bound(xs.begin(), xs.end(), x) - xs.begin()
#define ubv(xs, x) upper_bound(xs.begin(), xs.end(), x) - xs.begin()
#define bs(xs, x) binary_search(xs.begin(), xs.end(), x)

#define rep(i,n) for(auto i=0; i<(n); i++)
#define isValidPoint(x, y, mx, my) x >= 0 && x  < mx && y >= 0 && y < my

int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

const int MOD = 1000000007;

long exp_mod(long long a, long long b, long m) {
	long long res = 1;
	while (b > 0) {
		if (b&1)
			res = (res * a)%m;
		a = (a * a)%m;
		b >>= 1;
	}
	return (long)res;
}

long fact_mod(long n, long m) {
	long long res = 1;
	for (long i=n; i>0; --i)
		res = (res * i)%m;
	return (long)res;
}

long comb_mod(long n, long r) {
	long long res = 1;
	for (long i=0; i<r; ++i)
		res = (res * (n - i) )%MOD;
	return ( res * exp_mod(fact_mod(r, MOD), MOD-2, MOD) ) % MOD;
}

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

struct Pony
{
    int x,y;

    Pony(int x_, int y_)
    {
        x = x_;
        y = y_;
    }

    bool operator<(const Pony &another) const
    {
        return x < another.x;
    };
};

long modinv(long a, long m) {
    long b = m, u = 1, v = 0;
    while (b) {
        long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

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

  long n,x;
  cin >> n>>x;

  vector<long> l(n);
  fillLong(l, n);

  long cnt = 1;
  long d = 0;
  
  rep(i, n) {
    d = d + l[i];
    if (d <= x) cnt++;
    else break;
  }

  cout << cnt << endl;
}