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 nn. Initially the value of nnequals to vv and the player is able to do the following operation as many times as the player want (possibly zero): choose a positive integer xx that x<nx<n and xx is not a divisor of nn, then subtract xx from nn. The goal of the player is to minimize the value of nn 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: vv (1≤v≤1091≤v≤109), the initial value of nn.

Output

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

Examples

input

Copy

8

output

Copy

1

input

Copy

1

output

Copy

1

Note

In the first example, the player can choose x=3x=3 in the first turn, then nn becomes 55. He can then choose x=4x=4 in the second turn to get n=1n=1 as the result. There are other ways to get this minimum. However, for example, he cannot choose x=2x=2 in the first turn because 22 is a divisor of 88.

In the second example, since n=1n=1 initially, the player can do nothing.

題意：

給定一個數，每次可以減去任一個比其小的但不能是他的因子的數（也可以不減），輸出最後得到的最小值 。

分析：

任意相鄰的兩個數一定是互質的數，開始直接就輸出了1，wr，漏了一種情況就是當輸入的數是 2的時候，因為1是它的一個因子，不能減，所以其他的情況直接輸出1當為2時輸出2就可以了。

程式碼：

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(n==2) printf("2\n");
    else printf("1\n");
    return 0;
}