import java.util.*;

/**
 * The maximum number of trips is easy. Stamp all the pages consecutively 
 * until we cannot any more.
 * To get the minimum number of trips, try counting how many pages would
 * be filled if a gap of size p_i - 1 is left between each of the previously
 * stamped pages. To do that we just need to know how many pages were stamped
 * so far and how many trips we have taken. Use prefix sums.
 * Time complexity: O(n)
 *
 * @author Finn Lidbetter
 */

public class passport_finn {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long p = sc.nextLong();
        long[] vals = new long[n];
        for (int i=0; i<n; i++) {
            vals[i] = sc.nextLong();
        }
        int maxTrips = 0;
        long sum = 0;
        for (int i=0; i<n; i++) {
            sum += vals[i];
            if (sum<=p) {
                maxTrips = i+1;
            } else {
                break;
            }
        }
        long[] prefixSum = new long[n+1];
        for (int i=0; i<n; i++) {
            // This might overflow, but that's ok. We will detect
            // if prefixSum[i]>p and exit early later on. Since 2*p fits
            // within a long this will work ok.
            prefixSum[i+1] = prefixSum[i] + vals[i];
        }
        int minTrips = maxTrips;
        for (int i=0; i<n; i++) {
            if (prefixSum[i]>p || (vals[i]-1)>p/(i+1)) {
                minTrips = i;
                break;
            }
            long fill = prefixSum[i] + (i+1)*(vals[i]-1);
            if (fill>=p) {
                minTrips = i;
                break;
            }
        }
        // System.out.println(maxTrips);
        System.out.println(minTrips);
    }
}