//#pragma GCC optimize(2)
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <set>
#include <vector>
#include <utility>
#include <stack>
#include <deque>
#include <queue>
#include <string>
#include <cctype>
#include <sstream>
#include <map>
#include <random>
#include <unordered_map>
#include <unordered_set>
//#define LOCAL
#define ll long long
#define pdd pair<double, double>
#define pii pair<int, int>
#define pli pair<ll, int>
//#define mp(x, y) make_pair(x, y)
//#define ld long double
//#define int ll
#define maxn 200005
#define maxm 1000006
#define maxk 5004
#define maxprime 10000007
#define PI 3.14159265358979323846
#define mod 1000000007
#define eps 1e-6
#define INF 100000000000
using namespace std;

int gcd(int a, int b)
{
    return b==0 ? a : gcd(b, a%b);
}

int n;
int vis[maxn];

void gcd(ll a, ll b, ll &d, ll &x, ll &y)
{
    if(!b) { d = a; x = 1; y = 0; }
    else
    {
        gcd(b, a%b, d, y, x);
        y -= x*(a/b);
    }
}

ll inverse(ll a, ll p)
{
    ll d, x, y; a %= p;
    gcd(p, a, d, x, y);
    y = (y % p + p) % p;
    return y;
}


signed main()
{
#ifdef LOCAL
    freopen("data.in.txt","r",stdin);
    freopen("data.out.txt","w",stdout);
#endif
    cin >> n;
    int cnt = 0;
    for(int i = 1; i <= n; i++)
        if(gcd(n, i) == 1) vis[i] = 1, cnt++;
    ll mul = 1;
    for(int i = 1; i <= n; i++)
        if(vis[i]) mul = (mul * i) % n;
    mul = (mul+n) % n;
    if(mul == 1)
    {
        cout << cnt << endl;
        for(int i = 1; i <= n; i++)
            if(vis[i]) printf("%d ", i);
    }
    else
    {
        ll m = inverse(mul, n);
        vis[m] = 0;
        cout << cnt-1 << endl;
        for(int i = 1; i <= n; i++)
            if(vis[i]) printf("%d ", i);
    }
    cout << endl;
    return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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
//#pragma GCC optimize(2)
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <set>
#include <vector>
#include <utility>
#include <stack>
#include <deque>
#include <queue>
#include <string>
#include <cctype>
#include <sstream>
#include <map>
#include <random>
#include <unordered_map>
#include <unordered_set>
//#define LOCAL
#define ll long long
#define pdd pair<double, double>
#define pii pair<int, int>
#define pli pair<ll, int>
//#define mp(x, y) make_pair(x, y)
//#define ld long double
//#define int ll
#define maxn 200005
#define maxm 1000006
#define maxk 5004
#define maxprime 10000007
#define PI 3.14159265358979323846
#define mod 1000000007
#define eps 1e-6
#define INF 100000000000
using namespace std;

int gcd(int a, int b)
{
    return b==0 ? a : gcd(b, a%b);
}

int n;
int vis[maxn];

void gcd(ll a, ll b, ll &d, ll &x, ll &y)
{
    if(!b) { d = a; x = 1; y = 0; }
    else
    {
        gcd(b, a%b, d, y, x);
        y -= x*(a/b);
    }
}

ll inverse(ll a, ll p)
{
    ll d, x, y; a %= p;
    gcd(p, a, d, x, y);
    y = (y % p + p) % p;
    return y;
}


signed main()
{
#ifdef LOCAL
    freopen("data.in.txt","r",stdin);
    freopen("data.out.txt","w",stdout);
#endif
    cin >> n;
    int cnt = 0;
    for(int i = 1; i <= n; i++)
        if(gcd(n, i) == 1) vis[i] = 1, cnt++;
    ll mul = 1;
    for(int i = 1; i <= n; i++)
        if(vis[i]) mul = (mul * i) % n;
    mul = (mul+n) % n;
    if(mul == 1)
    {
        cout << cnt << endl;
        for(int i = 1; i <= n; i++)
            if(vis[i]) printf("%d ", i);
    }
    else
    {
        ll m = inverse(mul, n);
        vis[m] = 0;
        cout << cnt-1 << endl;
        for(int i = 1; i <= n; i++)
            if(vis[i]) printf("%d ", i);
    }
    cout << endl;
    return 0;
}
```