#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
void rd(long long &x) {
  x = 0;
  char s;
  s = getchar();
  long long f = 1;
  while (!isdigit(s)) {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (isdigit(s)) {
    x = x * 10 + s - '0';
    s = getchar();
  }
  x *= f;
  return;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void rep(long long &a, long long b) {
  a %= b;
  a += b;
  a %= b;
}
long long p;
long long mul(long long a, long long b) {
  long long res = 0;
  while (b) {
    if (b & 1) {
      res += a;
      rep(res, p);
    }
    b >>= 1;
    a = a + a;
    rep(a, p);
  }
  return res;
}
long long pow(long long a, long long b, long long p) {
  rep(a, p);
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a;
      rep(res, p);
    }
    b >>= 1;
    a = a * a;
    rep(a, p);
  }
  return res;
}
long long ex_gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = ex_gcd(b, a % b, x, y);
  long long z = x;
  x = y;
  y = z - a / b * y;
  return d;
}
long long bsgs(long long a, long long r, long long p) {
  map<long long, long long> ma;
  map<long long, bool> cunzai;
  ma.clear();
  cunzai.clear();
  long long tt = (long long)sqrt(p) + 1;
  for (long long i = 0; i < tt; i++) {
    long long val = pow(a, i, p) * r;
    rep(val, p);
    ma[val] = i;
    cunzai[val] = 1;
  }
  a = pow(a, tt, p);
  if (a == 0) {
    return r == 0 ? 1 : -1;
  }
  for (long long i = 0; i <= tt; i++) {
    long long tem = pow(a, i, p);
    if (cunzai[tem]) {
      long long j = ma[tem];
      if (j >= 0 and i * tt - j >= 0) return i * tt - j;
    }
  }
  return -1;
}
long long n;
long long euler(long long x) {
  long long tt = sqrt(x);
  long long res = x;
  for (long long i = 2; i <= tt; i++) {
    if (x % i == 0) {
      res = res / i * (i - 1);
      while (x % i == 0) {
        x /= i;
      }
    }
  }
  if (x > 1) {
    res = res / x * (x - 1);
  }
  return res;
}
bool isprime(long long x) {
  if (x == 1) return 0;
  if (x == 2) return 1;
  long long tt = sqrt(x);
  for (long long i = 2; i <= tt; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
bool isp[3 * 1123456];
long long sum[1123456];
void getprime() {
  memset(isp, 1, sizeof(isp));
  isp[1] = 0;
  for (long long i = 2; i < 3 * 1123456; i++) {
    if (isp[i] == 0) {
      continue;
    }
    for (long long j = i + i; j < 3 * 1123456; j += i) {
      isp[j] = 0;
    }
  }
}
signed main() {
  cin >> n;
  if (n == 4) {
    cout << "YES" << '\n';
    cout << 1 << '\n';
    cout << 3 << '\n';
    cout << 2 << '\n';
    cout << 4 << '\n';
    return 0;
  }
  if (n == 1) {
    cout << "YES" << '\n';
    cout << 1 << '\n';
    return 0;
  }
  if (!isprime(n)) {
    cout << "NO" << '\n';
    return 0;
  } else {
    cout << "YES" << '\n';
    for (long long i = 1; i <= n; i++) {
      if (i == 1) {
        cout << 1 << '\n';
      } else {
        long long x, y;
        long long d = ex_gcd(i - 1, n, x, y);
        x *= i / d;
        rep(x, n / d);
        if (x == 0) {
          x = n;
        }
        cout << x << '\n';
      }
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 mod = 1e9 + 7;
void rd(long long &x) {
  x = 0;
  char s;
  s = getchar();
  long long f = 1;
  while (!isdigit(s)) {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (isdigit(s)) {
    x = x * 10 + s - '0';
    s = getchar();
  }
  x *= f;
  return;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void rep(long long &a, long long b) {
  a %= b;
  a += b;
  a %= b;
}
long long p;
long long mul(long long a, long long b) {
  long long res = 0;
  while (b) {
    if (b & 1) {
      res += a;
      rep(res, p);
    }
    b >>= 1;
    a = a + a;
    rep(a, p);
  }
  return res;
}
long long pow(long long a, long long b, long long p) {
  rep(a, p);
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a;
      rep(res, p);
    }
    b >>= 1;
    a = a * a;
    rep(a, p);
  }
  return res;
}
long long ex_gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = ex_gcd(b, a % b, x, y);
  long long z = x;
  x = y;
  y = z - a / b * y;
  return d;
}
long long bsgs(long long a, long long r, long long p) {
  map<long long, long long> ma;
  map<long long, bool> cunzai;
  ma.clear();
  cunzai.clear();
  long long tt = (long long)sqrt(p) + 1;
  for (long long i = 0; i < tt; i++) {
    long long val = pow(a, i, p) * r;
    rep(val, p);
    ma[val] = i;
    cunzai[val] = 1;
  }
  a = pow(a, tt, p);
  if (a == 0) {
    return r == 0 ? 1 : -1;
  }
  for (long long i = 0; i <= tt; i++) {
    long long tem = pow(a, i, p);
    if (cunzai[tem]) {
      long long j = ma[tem];
      if (j >= 0 and i * tt - j >= 0) return i * tt - j;
    }
  }
  return -1;
}
long long n;
long long euler(long long x) {
  long long tt = sqrt(x);
  long long res = x;
  for (long long i = 2; i <= tt; i++) {
    if (x % i == 0) {
      res = res / i * (i - 1);
      while (x % i == 0) {
        x /= i;
      }
    }
  }
  if (x > 1) {
    res = res / x * (x - 1);
  }
  return res;
}
bool isprime(long long x) {
  if (x == 1) return 0;
  if (x == 2) return 1;
  long long tt = sqrt(x);
  for (long long i = 2; i <= tt; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
bool isp[3 * 1123456];
long long sum[1123456];
void getprime() {
  memset(isp, 1, sizeof(isp));
  isp[1] = 0;
  for (long long i = 2; i < 3 * 1123456; i++) {
    if (isp[i] == 0) {
      continue;
    }
    for (long long j = i + i; j < 3 * 1123456; j += i) {
      isp[j] = 0;
    }
  }
}
signed main() {
  cin >> n;
  if (n == 4) {
    cout << "YES" << '\n';
    cout << 1 << '\n';
    cout << 3 << '\n';
    cout << 2 << '\n';
    cout << 4 << '\n';
    return 0;
  }
  if (n == 1) {
    cout << "YES" << '\n';
    cout << 1 << '\n';
    return 0;
  }
  if (!isprime(n)) {
    cout << "NO" << '\n';
    return 0;
  } else {
    cout << "YES" << '\n';
    for (long long i = 1; i <= n; i++) {
      if (i == 1) {
        cout << 1 << '\n';
      } else {
        long long x, y;
        long long d = ex_gcd(i - 1, n, x, y);
        x *= i / d;
        rep(x, n / d);
        if (x == 0) {
          x = n;
        }
        cout << x << '\n';
      }
    }
  }
  return 0;
}

```