Some time ago Leonid have known about idempotent functions. Idempotent function defined on a set {1, 2, ..., n} is such function $\text{[math]}$ , that for any $\text{[math]}$ the formula g(g(x)) = g(x) holds.

Let's denote as f^(k)(x) the function f applied k times to the value x. More formally, f⁽¹⁾(x) = f(x), f^(k)(x) = f(f^(k - 1)(x)) for each k > 1.

You are given some function $\text{[math]}$ . Your task is to find minimum positive integer k such that function f^(k)(x) is idempotent.

Input

In the first line of the input there is a single integer n (1 ≤ n ≤ 200) — the size of function f domain.

In the second line follow f(1), f(2), ..., f(n) (1 ≤ f(i) ≤ n for each 1 ≤ i ≤ n), the values of a function.

Output

Output minimum k such that function f^(k)(x) is idempotent.

Note

In the first sample test function f(x) = f⁽¹⁾(x) is already idempotent since f(f(1)) = f(1) = 1, f(f(2)) = f(2) = 2, f(f(3)) = f(3) = 2, f(f(4)) = f(4) = 4.

In the second sample test:

function f(x) = f⁽¹⁾(x) isn't idempotent because f(f(1)) = 3 but f(1) = 2;
function f(x) = f⁽²⁾(x) is idempotent since for any x it is true that f⁽²⁾(x) = 3, so it is also true that f⁽²⁾(f⁽²⁾(x)) = 3.

In the third sample test:

function f(x) = f⁽¹⁾(x) isn't idempotent because f(f(1)) = 3 but f(1) = 2;
function f(f(x)) = f⁽²⁾(x) isn't idempotent because f⁽²⁾(f⁽²⁾(1)) = 2 but f⁽²⁾(1) = 3;
function f(f(f(x))) = f⁽³⁾(x) is idempotent since it is identity function: f⁽³⁾(x) = x for any $\text{[math]}$ meaning that the formula f⁽³⁾(f⁽³⁾(x)) = f⁽³⁾(x) also holds.

Idx	Name	Rating
A	Place Your Ad Here	2400
B	Duck Hunt	3100
C	Idempotent functions	2000
D	Superhero's Job	2600
E	Playing on Graph	2600
F	Quest	2100

VK Cup 2015 - Round 3 (unofficial online mirror, Div. 1 only)