Solve in Python:
Vasya and Petya wrote down all integers from 1 to n to play the "powers" game (n can be quite large; however, Vasya and Petya are not confused by this fact).

Players choose numbers in turn (Vasya chooses first). If some number x is chosen at the current turn, it is forbidden to choose x or all of its other positive integer powers (that is, x^2, x^3, ...) at the next turns. For instance, if the number 9 is chosen at the first turn, one cannot choose 9 or 81 later, while it is still allowed to choose 3 or 27. The one who cannot make a move loses.

Who wins if both Vasya and Petya play optimally?


-----Input-----

Input contains single integer n (1 ≤ n ≤ 10^9).


-----Output-----

Print the name of the winner — "Vasya" or "Petya" (without quotes).


-----Examples-----
Input
1

Output
Vasya

Input
2

Output
Petya

Input
8

Output
Petya



-----Note-----

In the first sample Vasya will choose 1 and win immediately.

In the second sample no matter which number Vasya chooses during his first turn, Petya can choose the remaining number and win.
from sys import stdin, stdout
import math, collections
mod = 10**9+7

def isPower(n):
    if (n <= 1):
        return True
    for x in range(2, (int)(math.sqrt(n)) + 1):
        p = x
        while (p <= n):
            p = p * x
            if (p == n):
                return True

    return False
n = int(input())
arr = [0,1,2,1,4,3,2,1,5,6,2,1,8,7,5,9,8,7,3,4,7,4,2,1,10,9,3,6,11,12]
ans = arr[int(math.log(n, 2))]
s = int(math.log(n, 2))
for i in range(3, int(n**0.5)+1):
    if not isPower(i):
        ans^=arr[int(math.log(n, i))]
        s+=int(math.log(n, i))
ans^=((n-s)%2)
print("Vasya" if ans else "Petya")