#include <bits/stdc++.h>
using namespace std;
long long kpow(long long x, int y, int p) {
  if (y == 0) return 1;
  long long w = kpow(x, y / 2, p);
  w = w * w;
  w %= p;
  if (y % 2) w *= x;
  w %= p;
  return w;
}
int n;
int main() {
  scanf("%d", &n);
  if (n == 4) {
    cout << "YES" << endl;
    cout << 1 << endl;
    cout << 3 << endl;
    cout << 2 << endl;
    cout << 4 << endl;
    return 0;
  }
  int bj = 0;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      cout << "NO";
      return 0;
    }
  }
  printf("YES\n");
  printf("1\n");
  for (int i = 2; i < n; i++) {
    long long ans = 1;
    ans *= i;
    ans *= kpow(i - 1, n - 2, n);
    ans %= n;
    printf("%I64d\n", ans);
  }
  if (n != 1) printf("%d", n);
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

Consider a sequence [a1, a2, ... , an]. Define its prefix product sequence <image>.

Now given n, find a permutation of [1, 2, ..., n], such that its prefix product sequence is a permutation of [0, 1, ..., n - 1].

Input

The only input line contains an integer n (1 ≤ n ≤ 105).

Output

In the first output line, print "YES" if such sequence exists, or print "NO" if no such sequence exists.

If any solution exists, you should output n more lines. i-th line contains only an integer ai. The elements of the sequence should be different positive integers no larger than n.

If there are multiple solutions, you are allowed to print any of them.

Examples

Input

7


Output

YES
1
4
3
6
5
2
7


Input

6


Output

NO

Note

For the second sample, there are no valid sequences.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long kpow(long long x, int y, int p) {
  if (y == 0) return 1;
  long long w = kpow(x, y / 2, p);
  w = w * w;
  w %= p;
  if (y % 2) w *= x;
  w %= p;
  return w;
}
int n;
int main() {
  scanf("%d", &n);
  if (n == 4) {
    cout << "YES" << endl;
    cout << 1 << endl;
    cout << 3 << endl;
    cout << 2 << endl;
    cout << 4 << endl;
    return 0;
  }
  int bj = 0;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      cout << "NO";
      return 0;
    }
  }
  printf("YES\n");
  printf("1\n");
  for (int i = 2; i < n; i++) {
    long long ans = 1;
    ans *= i;
    ans *= kpow(i - 1, n - 2, n);
    ans %= n;
    printf("%I64d\n", ans);
  }
  if (n != 1) printf("%d", n);
  return 0;
}

```