#include <bits/stdc++.h>
using namespace std;
constexpr int MOD = 1000000007;
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
int logceil(long long first) {
  int b = 0;
  while (first) {
    first >>= 1;
    ++b;
  }
  return b;
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.first) ^ (u(p.second) << 7);
  }
};
}  // namespace std
template <typename T, typename F>
T bsh(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      l = m + 1;
      r = m;
    } else {
      h = m - 1;
    }
  }
  return r;
}
template <typename T, typename F>
T bsl(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      h = m - 1;
      r = m;
    } else {
      l = m + 1;
    }
  }
  return r;
}
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector<vector<T>>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector<vector<T>>>(a, vector<vector<T>>(b, vector<T>(c, t))) {}
};
class TaskC {
 public:
  void solve(istream& cin, ostream& cout) {
    int N;
    cin >> N;
    if (N == 1) {
      cout << "YES\n1\n";
      return;
    }
    if (N == 4) {
      cout << "YES\n1\n3\n2\n4";
      return;
    }
    for (int i = 2; i * i <= N; ++i) {
      if (N % i == 0) {
        cout << "NO\n";
        return;
      }
    }
    auto inv = [&](int a) {
      unsigned int r = 1, e = a, p = N - 2;
      while (p) {
        if (p & 1) {
          r = ((long long)r * e) % N;
        }
        e = ((long long)e * e) % N;
        p >>= 1;
      }
      return r;
    };
    cout << "YES\n1\n";
    for (int i = 2; i < N; ++i) {
      cout << ((long long)i * inv(i - 1)) % N << '\n';
    }
    cout << N << '\n';
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  TaskC solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
constexpr int MOD = 1000000007;
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
int logceil(long long first) {
  int b = 0;
  while (first) {
    first >>= 1;
    ++b;
  }
  return b;
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.first) ^ (u(p.second) << 7);
  }
};
}  // namespace std
template <typename T, typename F>
T bsh(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      l = m + 1;
      r = m;
    } else {
      h = m - 1;
    }
  }
  return r;
}
template <typename T, typename F>
T bsl(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      h = m - 1;
      r = m;
    } else {
      l = m + 1;
    }
  }
  return r;
}
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector<vector<T>>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector<vector<T>>>(a, vector<vector<T>>(b, vector<T>(c, t))) {}
};
class TaskC {
 public:
  void solve(istream& cin, ostream& cout) {
    int N;
    cin >> N;
    if (N == 1) {
      cout << "YES\n1\n";
      return;
    }
    if (N == 4) {
      cout << "YES\n1\n3\n2\n4";
      return;
    }
    for (int i = 2; i * i <= N; ++i) {
      if (N % i == 0) {
        cout << "NO\n";
        return;
      }
    }
    auto inv = [&](int a) {
      unsigned int r = 1, e = a, p = N - 2;
      while (p) {
        if (p & 1) {
          r = ((long long)r * e) % N;
        }
        e = ((long long)e * e) % N;
        p >>= 1;
      }
      return r;
    };
    cout << "YES\n1\n";
    for (int i = 2; i < N; ++i) {
      cout << ((long long)i * inv(i - 1)) % N << '\n';
    }
    cout << N << '\n';
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  TaskC solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

```