题目背景

Abusing judging resources leads to an account ban.

You can self-test the first three test cases of each Subtask at the link: https://www.luogu.com.cn/problem/U576602 to avoid long judging wait times.

题目描述

Since the CCO often uses integers, Alice needs to learn about the integers! A positive integer $n$ can be written in base $b$ as the sequence $d_{m-1}d_{m-2} \ldots d_{1}d_{0}$ if the following hold:

• Each digit $d_{i}$ is between 0 and $b-1$, inclusive.
• $d_{m-1}>0$.
• $n=d_{m-1} \times b^{m-1}+d_{m-2} \times b^{m-2}+\cdots+d_{1} \times b^{1}+d_{0} \times b^{0}$.

For example, the integer 2025 in base 19 is the sequence $(5,11,11)$ because $2025=5 \times 19^{2}+11 \times 19^{1}+11 \times 19^{0}$.

An integer $n$ is $b$-balanced if, when $n$ is written in base $b$, the average of the digits is $\frac{b-1}{2}$. For example, 2025 is 19-balanced because $\frac{5+11+11}{3}=9=\frac{19-1}{2}$.

Alice can easily find integers that are 19-balanced. However, she has trouble finding integers that are balanced in multiple ways. Given $B$ and $N$, please help Alice find the minimum integer $x$ such that:

• $x$ is $b$-balanced, for all $2 \leq b \leq B$.
• $x \geq N$.

输入格式

The first line of input contains two space-separated integers $B$ and $N$ ($N \geq 1$).

It is guaranteed that the answer does not exceed $10^{18}$.

输出格式

Output the minimum integer $x$ from the problem statement.

4 100

141

7 10000000000

16926961207710

提示

Sample 1 Explanation

$141$ in base $2$ is $10001101$. The average digit is

$$\frac{1+0+0+0+1+1+0+1}{8}=0.5=\frac{2-1}{2}.$$

Therefore, $141$ is 2-balanced.

$141$ in base 3 is $12020$. The average digit is

$$\frac{1+2+0+2+0}{5}=1=\frac{3-1}{2}.$$

Therefore, $141$ is 3-balanced.

$141$ in base 4 is $2031$. The average digit is

$$\frac{2+0+3+1}{4}=1.5=\frac{4-1}{2}.$$

Therefore, $141$ is 4-balanced.

Lastly, $141 \geq 100$.

Feel free to use these code snippets as part of your solution.

// Important: If x is 0, the result is undefined.
int base_2_length(unsigned long long x) {
  return 64-__builtin_clzll(x);
}

int base_2_sum(unsigned long long x) {
  return __builtin_popcountll(x);
}

The following table shows how the available $25$ marks are distributed:

Marks Awarded	Bounds on $B$	Bounds on $N$
2 marks	$2 \leq B \leq 7$	$1 \leq N \leq 10^{4}$
6 marks	$2 \leq B \leq 6$	$N = 10^{10}$
2 marks	$2 \leq B \leq 7$	None
9 marks	$8 \leq B \leq 11$	$N = 1$
4 marks	$B = 8$	None
2 marks	$9 \leq B \leq 11$