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 <iomanip>
#include <cstdio>
#include <cstring>
#include <string>
#include <sstream>
#include <algorithm>
#include <utility>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <cmath>

#define mst(a,b)memset(a, b, sizeof(a))
#define repi(a,b)for(int i = a; i < b; i++)
#define repj(a,b)for(int j = a; j < b; j++)
#define pii pair<int,int>

#ifdef LOCAL
#define fre()freopen("in.txt","r",stdin)
#endif
#ifndef LOCAL
#define fre()
#endif

//static int x = []() {
//	std::ios::sync_with_stdio(false);
//	std::cin.tie(NULL);
//	return 0;
//}();

#define FASTIO
#ifdef FASTIO
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
//#define scanf abcabc
//#define printf abcabc
#endif
//#ifndef FASTIO
//#define cin abcabc
//#define cout abcabc
//#endif

typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;

using namespace std;


int main() {
	int n, x, ans = 1, sum = 0; cin >> n >> x;
	repi(0, n) {
		int tmp; cin >> tmp;
		sum += tmp;
		if (sum <= x)ans++;
		else break;
	}
	cout << ans << endl;
	return 0;
}