Chouti was doing a competitive programming competition. However, after having all the problems accepted, he got bored and decided to invent some small games.

He came up with the following game. The player has a positive integer n. Initially the value of n equals to v and the player is able to do the following operation as many times as the player want (possibly zero): choose a positive integer x that x<n and x is not a divisor of n, then subtract x from n. The goal of the player is to minimize the value of n in the end.

Soon, Chouti found the game trivial. Can you also beat the game?

Input

The input contains only one integer in the first line: v (1 ≤ v ≤ 10^9), the initial value of n.

Output

Output a single integer, the minimum value of n the player can get.

Examples

Input


8


Output


1


Input


1


Output


1

Note

In the first example, the player can choose x=3 in the first turn, then n becomes 5. He can then choose x=4 in the second turn to get n=1 as the result. There are other ways to get this minimum. However, for example, he cannot choose x=2 in the first turn because 2 is a divisor of 8.

In the second example, since n=1 initially, the player can do nothing.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 1000000007;
double eps = 1e-12;
long long getSum(string str) {
  long long sum = 0;
  for (int i = 0; i < str.length(); i++) {
    sum = sum + str[i] - 48;
  }
  return sum;
}
double round_up(double value, int decimal_places) {
  const double multiplier = std::pow(10.0, decimal_places);
  return std::ceil(value * multiplier) / multiplier;
}
bool isSquare(long long n) {
  long long res = sqrt(n);
  return res * res == n;
}
void solve() {
  long long n;
  cin >> n;
  if (n == 2)
    cout << 2 << "\n";
  else
    cout << 1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (int it = 1; it <= t; it++) {
    solve();
  }
  return 0;
}
