#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include<time.h>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <stdio.h>
#include <deque>
#include <map>
#include <set>
#include <string>
#define ll long long 
typedef long long LL;
typedef unsigned long long ULL;
//typedef long double ld;
using namespace std;
const int MAXN = 2e6 + 5000;
const long double eps = (1e-12);
const int logn = 21;
const double Pi = acos(-1.0);
const LL pp = 1000000000 + 7;
int lowbit(int x) { return x & (-x); }
#define MP(i,j) make_pair(i,j)
#define M(x) ((((x)%pp)+pp)%pp)
#define int LL
#define P pair<int,int>
const int INF = 1e9;

int gcd(int a, int b) {
	if (b == 0) return a;
	else return gcd(b, a % b);
}

vector<int> ans;
int n;

bool check(int n) {
	ans.clear();
	int tmp = 1;
	for (int i = 1; i < n; i++) {
		if (gcd(i, n) == 1) {
			ans.push_back(i);
			tmp *= i;
			tmp %= n;
		}
	}
	if (tmp == 1) return true;
	for (auto a : ans) {
		if (a == tmp) return true;
	}
	return false;
}

signed main() {
	//int t = 1000;
	//while (t--) {
	//	if (check(t) == false) {
	//		cout << t << endl;
	//	}
	//}
	scanf("%lld", &n);

	int tmp = 1;
	for (int i = 1; i < n; i++) {
		if (gcd(n, i) == 1) {
			ans.push_back(i);
			tmp *= i;
			tmp %= n;
		}
	}

	int sz = ans.size();
	if (tmp == 1) {
		printf("%d\n", sz);
		for (auto a : ans) {
			printf("%d ", a);
		}
	}
	else {
		sz--;
		printf("%d\n", sz);
		for (auto a : ans) {
			if (a != tmp) {
				printf("%d ", a);
			}
		}
	}
	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
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include<time.h>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <stdio.h>
#include <deque>
#include <map>
#include <set>
#include <string>
#define ll long long 
typedef long long LL;
typedef unsigned long long ULL;
//typedef long double ld;
using namespace std;
const int MAXN = 2e6 + 5000;
const long double eps = (1e-12);
const int logn = 21;
const double Pi = acos(-1.0);
const LL pp = 1000000000 + 7;
int lowbit(int x) { return x & (-x); }
#define MP(i,j) make_pair(i,j)
#define M(x) ((((x)%pp)+pp)%pp)
#define int LL
#define P pair<int,int>
const int INF = 1e9;

int gcd(int a, int b) {
	if (b == 0) return a;
	else return gcd(b, a % b);
}

vector<int> ans;
int n;

bool check(int n) {
	ans.clear();
	int tmp = 1;
	for (int i = 1; i < n; i++) {
		if (gcd(i, n) == 1) {
			ans.push_back(i);
			tmp *= i;
			tmp %= n;
		}
	}
	if (tmp == 1) return true;
	for (auto a : ans) {
		if (a == tmp) return true;
	}
	return false;
}

signed main() {
	//int t = 1000;
	//while (t--) {
	//	if (check(t) == false) {
	//		cout << t << endl;
	//	}
	//}
	scanf("%lld", &n);

	int tmp = 1;
	for (int i = 1; i < n; i++) {
		if (gcd(n, i) == 1) {
			ans.push_back(i);
			tmp *= i;
			tmp %= n;
		}
	}

	int sz = ans.size();
	if (tmp == 1) {
		printf("%d\n", sz);
		for (auto a : ans) {
			printf("%d ", a);
		}
	}
	else {
		sz--;
		printf("%d\n", sz);
		for (auto a : ans) {
			if (a != tmp) {
				printf("%d ", a);
			}
		}
	}
	return 0;
}
```