Problem: Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:

gcd(sum(S_1), sum(S_2)) > 1 

Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).

Every integer number from 1 to n should be present in exactly one of S_1 or S_2.

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)

Output

If such partition doesn't exist, print "No" (quotes for clarity).

Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.

Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.

If there are multiple possible partitions — print any of them.

Examples

Input

1


Output

No

Input

3


Output

Yes
1 2
2 1 3 

Note

In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".

In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) {
  return n | (1LL << i);
  ;
}
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline bool isIntege(double num) { return (num == (int)num); }
inline bool collinear(int x1, int y1, int x2, int y2, int x3, int y3) {
  return (y1 - y2) * (x1 - x3) == (y1 - y3) * (x1 - x2);
}
inline double coDist(double x1, double y1, double x2, double y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
struct edge {
  int u, v, w;
};
bool cmp(edge &a, edge &b) { return a.w < b.w; }
string to_s(int t) {
  stringstream ss;
  ss << t;
  return ss.str();
}
bool seive[1010000];
vector<int> prime;
void seiveGen(int limit) {
  limit += 100;
  int sqrtn = sqrt(limit);
  for (int i = 3; i <= sqrtn; i += 2) {
    if (!seive[i >> 1]) {
      for (int j = i * i; j < limit; j += i + i) {
        seive[j >> 1] = 1;
      }
    }
  }
  prime.push_back(2);
  for (int i = 3; i < limit; i += 2) {
    if (!seive[i >> 1]) prime.push_back(i);
  }
}
vector<int> PrimeF;
void primeFactorize(int n) {
  for (int i = 0; prime[i] * prime[i] <= n; i++) {
    if (n % prime[i] == 0) {
      while (n % prime[i] == 0) {
        n /= prime[i];
        PrimeF.push_back(prime[i]);
      }
    }
  }
  if (n > 1) {
    PrimeF.push_back(n);
  }
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  long long s1, s2, x = 0;
  s1 = (n * (n + 1)) / 2;
  seiveGen(sqrt(s1 + 10000));
  primeFactorize(s1);
  if (n <= 2) {
    return cout << "No" << '\n', 0;
  } else {
    sort(PrimeF.begin(), PrimeF.end());
    for (int i = 0; i < PrimeF.size(); i++) {
      if (PrimeF[i] != n) {
        s2 = PrimeF[i];
        x = 1;
        break;
      }
    }
    if (x == 0) return cout << "No" << '\n', 0;
    cout << "Yes" << '\n';
    cout << 1 << " " << s2 << '\n';
    cout << n - 1 << " ";
    for (int i = 1; i <= n; i++) {
      if (i != s2) cout << i << " ";
    }
  }
}