#include <bits/stdc++.h>
using namespace std;
const long long inf = 2147383647;
const double pi = 2 * acos(0.0);
const double eps = 1e-9;
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
inline long long gcd(long long a, long long b) {
  a = ((a) < 0 ? -(a) : (a));
  b = ((b) < 0 ? -(b) : (b));
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
int ext_gcd(int a, int b, int& x, int& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  int x1, y1;
  int d = ext_gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
inline long long modInv(int a, int m) {
  int x, y;
  ext_gcd(a, m, x, y);
  if (x < 0) x += m;
  return x;
}
inline long long power(long long a, long long p) {
  long long res = 1, x = a;
  while (p) {
    if (p & 1) res = (res * x);
    x = (x * x);
    p >>= 1;
  }
  return res;
}
inline long long bigmod(long long a, long long p, long long m) {
  long long res = 1, x = a % m;
  while (p) {
    if (p & 1) res = (res * x) % m;
    x = (x * x) % m;
    p >>= 1;
  }
  return res;
}
bool isPrime(int n) {
  int sqrtn = sqrt(n);
  for (int i = (2); i <= (sqrtn); ++i) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  int n;
  scanf("%d", &n);
  if (n == 4) {
    printf("YES\n");
    printf("%d\n%d\n%d\n%d\n", 1, 3, 2, 4);
    return 0;
  }
  if (n == 1) {
    printf("YES\n");
    printf("1\n");
    return 0;
  }
  if (isPrime(n) == false) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  printf("1\n");
  for (int i = (2); i <= (n - 1); ++i) {
    printf("%d\n", (i * bigmod(i - 1, n - 2, n)) % n);
  }
  printf("%d\n", n);
  return 0;
}

### Prompt

Your task is to create 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;
const long long inf = 2147383647;
const double pi = 2 * acos(0.0);
const double eps = 1e-9;
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
inline long long gcd(long long a, long long b) {
  a = ((a) < 0 ? -(a) : (a));
  b = ((b) < 0 ? -(b) : (b));
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
int ext_gcd(int a, int b, int& x, int& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  int x1, y1;
  int d = ext_gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
inline long long modInv(int a, int m) {
  int x, y;
  ext_gcd(a, m, x, y);
  if (x < 0) x += m;
  return x;
}
inline long long power(long long a, long long p) {
  long long res = 1, x = a;
  while (p) {
    if (p & 1) res = (res * x);
    x = (x * x);
    p >>= 1;
  }
  return res;
}
inline long long bigmod(long long a, long long p, long long m) {
  long long res = 1, x = a % m;
  while (p) {
    if (p & 1) res = (res * x) % m;
    x = (x * x) % m;
    p >>= 1;
  }
  return res;
}
bool isPrime(int n) {
  int sqrtn = sqrt(n);
  for (int i = (2); i <= (sqrtn); ++i) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  int n;
  scanf("%d", &n);
  if (n == 4) {
    printf("YES\n");
    printf("%d\n%d\n%d\n%d\n", 1, 3, 2, 4);
    return 0;
  }
  if (n == 1) {
    printf("YES\n");
    printf("1\n");
    return 0;
  }
  if (isPrime(n) == false) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  printf("1\n");
  for (int i = (2); i <= (n - 1); ++i) {
    printf("%d\n", (i * bigmod(i - 1, n - 2, n)) % n);
  }
  printf("%d\n", n);
  return 0;
}

```