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, x2, x3, ...) 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 ≤ 109).

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.
#include <bits/stdc++.h>
using namespace std;
const int sq = 40 * 1000;
bool used[sq + 1];
const int G[30] = {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};
int main() {
  int n;
  cin >> n;
  int res = 0;
  int after_sq = 0;
  for (int i = 1; i <= sq; ++i) {
    if (used[i]) {
      continue;
    }
    int cnt = i == 1;
    for (long long j = i; i != 1 && j <= n; j *= i) {
      if (j <= sq) {
        used[j] = true;
      } else {
        ++after_sq;
      }
      ++cnt;
    }
    res ^= G[cnt];
  }
  res ^= (n > sq ? (n - sq - after_sq) : 0) & 1;
  if (res) {
    cout << "Vasya\n";
  } else {
    cout << "Petya\n";
  }
  return 0;
}
