#include<bits/stdc++.h>
#define sf(x) scanf("%d",&x)
#define pf(x) printf("%d ",x)
#define sf2(x,y) scanf("%d %d",&x,&y)
#define pf2(x,y) printf("%d %d\n",x,y)
#define sf3(x,y,z) scanf("%d %d %d",&x,&y,&z)
#define pf3(x,y,z) printf("%d %d %d\n",x,y,z)
#define sfc(c) scanf(" %c",&c)
#define pfc(c) printf("%c",c)
#define pb(x) push_back(x)
#define sfl(x) scanf("%lld",&x)
#define mp(x,y) make_pair(x,y)
#define INF 1000000008
#define LLINF ll(1e18+10)
#define ENDL puts("")

#ifdef DBG
#define dbg(x) cout<<'\t'<<#x<<' '<<x<<endl
#else
#define dbg(x)
#endif

#define forn(i,n) for(int i=0; i<n; i++)
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef unsigned int uint;
typedef long double ld;
 
 
template<class T>
void printVector(T begin, T end)
{
    for(auto i=begin; i!=end; i++)
    {
        cout<<*i<<' ';
    }
    cout<<endl;
}



int gcd (int a, int b, int & x, int & y) {
	if (a == 0) {
		x = 0; y = 1;
		return b;
	}
	int x1, y1;
	int d = gcd (b%a, a, x1, y1);
	x = y1 - (b / a) * x1;
	y = x1;
	return d;
}





int main()
{
    /*

    /**/
    

    int n;
    sf(n);
    set<int> res;
    int x,y;
    for(int i=1; i<n; i++)
    {
        x=y=0;
        int g=gcd(i,n,x,y);
        if(g==1)
        {
            res.insert(i);
        }
    }
    ll m=1;
    for(auto i: res)
    {
        m=(m*i)%n;
    }
    if(m!=1)
    {
        res.erase(m);
    }
    printf("%d\n",res.size());
    for(auto i: res)
    {
        printf("%d ",i);
    }
    
    
    
    
    return 0;
 
}
 
 
 
 
 
 
 
 
 
### Prompt

Construct a Cpp code solution to the problem outlined:

Now you get Baby Ehab's first words: "Given an integer n, find the longest subsequence of [1,2, …, n-1] whose product is 1 modulo n." Please solve the problem.

A sequence b is a subsequence of an array a if b can be obtained from a by deleting some (possibly all) elements. The product of an empty subsequence is equal to 1.

Input

The only line contains the integer n (2 ≤ n ≤ 10^5).

Output

The first line should contain a single integer, the length of the longest subsequence.

The second line should contain the elements of the subsequence, in increasing order.

If there are multiple solutions, you can print any.

Examples

Input


5


Output


3
1 2 3 

Input


8


Output


4
1 3 5 7 

Note

In the first example, the product of the elements is 6 which is congruent to 1 modulo 5. The only longer subsequence is [1,2,3,4]. Its product is 24 which is congruent to 4 modulo 5. Hence, the answer is [1,2,3].

### Response

```cpp
#include<bits/stdc++.h>
#define sf(x) scanf("%d",&x)
#define pf(x) printf("%d ",x)
#define sf2(x,y) scanf("%d %d",&x,&y)
#define pf2(x,y) printf("%d %d\n",x,y)
#define sf3(x,y,z) scanf("%d %d %d",&x,&y,&z)
#define pf3(x,y,z) printf("%d %d %d\n",x,y,z)
#define sfc(c) scanf(" %c",&c)
#define pfc(c) printf("%c",c)
#define pb(x) push_back(x)
#define sfl(x) scanf("%lld",&x)
#define mp(x,y) make_pair(x,y)
#define INF 1000000008
#define LLINF ll(1e18+10)
#define ENDL puts("")

#ifdef DBG
#define dbg(x) cout<<'\t'<<#x<<' '<<x<<endl
#else
#define dbg(x)
#endif

#define forn(i,n) for(int i=0; i<n; i++)
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef unsigned int uint;
typedef long double ld;
 
 
template<class T>
void printVector(T begin, T end)
{
    for(auto i=begin; i!=end; i++)
    {
        cout<<*i<<' ';
    }
    cout<<endl;
}



int gcd (int a, int b, int & x, int & y) {
	if (a == 0) {
		x = 0; y = 1;
		return b;
	}
	int x1, y1;
	int d = gcd (b%a, a, x1, y1);
	x = y1 - (b / a) * x1;
	y = x1;
	return d;
}





int main()
{
    /*

    /**/
    

    int n;
    sf(n);
    set<int> res;
    int x,y;
    for(int i=1; i<n; i++)
    {
        x=y=0;
        int g=gcd(i,n,x,y);
        if(g==1)
        {
            res.insert(i);
        }
    }
    ll m=1;
    for(auto i: res)
    {
        m=(m*i)%n;
    }
    if(m!=1)
    {
        res.erase(m);
    }
    printf("%d\n",res.size());
    for(auto i: res)
    {
        printf("%d ",i);
    }
    
    
    
    
    return 0;
 
}
 
 
 
 
 
 
 
 
 
```