Labs_test1 A - Fedya and Maths 欧拉定理

Labs_test1  A - Fedya and Maths  欧拉定理

A - Fedya and Maths

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:

(1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ) mod 5

for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e.g. it can exceed any integer type of your programming language).

Input

The single line contains a single integer n (0 ≤ n ≤ 10¹⁰⁵). The number doesn't contain any leading zeroes.

Output

Print the value of the expression without leading zeros.

Sample Input

Input

4

Output

4

Input

124356983594583453458888889

Output

0

Hint

Operation x mod y means taking remainder after division x by y.

Note to the first sample:

思路：由欧拉定理：a,n互质，则a^ph(n)=1 (mod n)。

         ∑ i^n mod 5 = ∑ i^(n/ph(5) *ph(5) +n mod ph(5)) mod n

                            = ∑ (i^(k*ph(5))*(i^n mod ph(5)) mod n

                            = ∑ 1 * (i^ (n mod ph(5))mod n

以及用同余模求大数n mod ph(5)

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转载于:https://www.cnblogs.com/--560/p/4688061.html