import java.util.Scanner;

/**
 * EggDropping
 * You are given k identical eggs and you have access to a building with n
 * floors labeled from 1 to n.
 * 
 * You know that there exists a floor f where 0 <= f <= n such that any egg
 * dropped at a floor higher than f will break,
 * and any egg dropped at or below floor f will not break.
 * 
 * Each move, you may take an unbroken egg and drop it from any floor x (where 1
 * <= x <= n).
 * If the egg breaks, you can no longer use it. However, if the egg does not
 * break, you may reuse it in future moves.
 * 
 * Return the minimum number of moves that you need to determine with certainty
 * what the value of f is.
 * 
 * Example 1:
 * 
 * Input: k = 1, n = 2
 * Output: 2
 * Explanation:
 * Drop the egg from floor 1. If it breaks, we know that f = 0.
 * Otherwise, drop the egg from floor 2. If it breaks, we know that f = 1.
 * If it does not break, then we know f = 2.
 * Hence, we need at minimum 2 moves to determine with certainty what the value
 * of f is.
 * 
 */
public class EggDropping {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int a = scanner.nextInt();
    int b = scanner.nextInt();
    scanner.close();
    System.out.println(minMoves(a, b));
  }

  private static int minMoves(int k, int n) {
    // if n > f ==> eggs break
    // if n < f ==> eggs are good
    int min = 0;
    int[][] dp = new int[n + 1][k + 1];
    while (dp[min][k] < n) {
      min++;
      for (int i = 1; i <= k; i++) {
        dp[min][i] = dp[min - 1][i - 1] + dp[min - 1][i] + 1;
      }
    }
    return min;
  }
}