Strange Limit#5665

Strange Limit

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Very easy

Consider the sequence $a_{n}$ a_n defined by the following recurrence:

a_{1} = p, a_{n + 1} = p^{a_{n}} for n \geq 1,

a_1 = p,\\ a_{n+1} = p^{a_n}\quad \text{for } n \ge 1,

where $p$ is a prime number.

Let

b_{n} = a_{n} mod m!,

b_n = a_n \bmod m!,

where $m!$ denotes the factorial of $m$ , that is, $m! = 1 \times 2 \times \dots \times m$ m! = 1 \times 2 \times \dots \times m.

It may seem strange, but for all $p$ and all $m$ , the sequence $b_{n}$ b_n has a limit as $n \to + \infty$ n \to +\infty. Your task is to find this limit. Given $p$ and $m$ , find this value:

n \to + \infty lim (a_{n} mod m!) .

\lim_{n \to +\infty} (a_n \bmod m!).

Input

The input contains two integers $p$ and $m$ ( $2 \leq p$ 2 \le p, $m \leq 12$ m \le 12, and $p$ is prime).

Output

Output the limit described above.

Examples

Example 1

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a_{n}

a_{1} = p, a_{n + 1} = p^{a_{n}} for n \geq 1,

b_{n} = a_{n} mod m!,

m! = 1 \times 2 \times \dots \times m

b_{n}

n \to + \infty

n \to + \infty lim (a_{n} mod m!) .

2 \leq p

m \leq 12