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: Wrong Answer)

#include <bits/stdc++.h>
#define FOR(i, a, n) for(int (i) = (int)(a); (i) < (int)(n); ++(i))
#define FORe(i, a, n) for(int (i) = (int)(a); (i) <= (int)(n); ++(i))
#define rFOR(i, a, n) for(int (i) = (int)(n)-1; (i) >= (int)(a); --(i))
#define rFORe(i, a, n) for(int (i) = (int)(n); (i) >= (int)(a); --(i))
#define REP(i, n) FOR((i), 0, (n))
#define REPe(i, n) FORe((i), 1, (n))
#define REPo(i, n) FORe((i), 0, (n))
#define rREP(i, n) rFOR((i), 0, (n))
#define rREPe(i, n) rFORe((i), 1, (n))
#define rREPo(i, n) rFORe((i), 0, (n))
#define rep(i, n) FOR((i), 0, (n))
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define fi first
#define se second
#define dup(x, y) (((x)+(y)-1)/(y))
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
using ll = long long;
using P = pair<int,int>;
using T = tuple<int,int,int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vs = vector<string>;
using vp = vector<P>;
using vt = vector<T>;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main() {
  int n, x;
  cin >> n >> x;
  vi l(n);
  REP(i, n) cin >> l[i];
  int d = 0;
  int ans = n;
  REP(i, n) {
    if (d > x) {
      ans = i;
      break;
    }
    d += l[i];
  }
  cout << ans << endl;

  return 0;
}