#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 200010;
int a[N];
ll pre[N]; 
int n, S;

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

    cin >> n >> S;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        pre[i] = pre[i-1] + a[i]; 
    }

    int l = 1, ans = 0;
    // 枚举右端点r，找最左的l使得 [l, r] 的和 ≤ S
    for (int r = 1; r <= n; ++r) {
        while (pre[r] - pre[l-1] > S) {
            l++;
        }
        ans = max(ans, r - l + 1);
    }

    cout << ans << endl;
    return 0;
}