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>
using namespace std;
int getFldsFromLine(char *line, std::vector<char *> &res,
                    const char *sep = " \t\n\v\f\r");
int main(int argc, char **argv) {
  FILE *infile = stdin;
  char *line;
  line = (char *)calloc(5000000, sizeof(char));
  if (line == NULL) {
    fprintf(stderr, "Couldn't allocate space for '%s'\nBye!\n", "line");
    exit(-1);
  };
  std::vector<char *> flds;
  int numFlds, numCases, caseNum = 0;
  {
    char *feabeqav = fgets(line, 5000000, infile);
    numFlds = getFldsFromLine(line, flds);
  };
  int k = atoi(flds[0]);
  if (k != 2) k = 1;
  cout << k << endl;
  return (0);
  return (0);
}
int getFldsFromLine(char *line, std::vector<char *> &res, const char *sep) {
  res.clear();
  char *tok = strtok(line, sep);
  while (tok) {
    res.push_back(tok);
    tok = strtok(0, sep);
  }
  return res.size();
}
