Problem: Seyyed and MoJaK are friends of Sajjad. Sajjad likes a permutation. Seyyed wants to change the permutation in a way that Sajjad won't like it. Seyyed thinks more swaps yield more probability to do that, so he makes MoJaK to perform a swap between every pair of positions (i, j), where i < j, exactly once. MoJaK doesn't like to upset Sajjad.

Given the permutation, determine whether it is possible to swap all pairs of positions so that the permutation stays the same. If it is possible find how to do that. 

Input

The first line contains single integer n (1 ≤ n ≤ 1000) — the size of the permutation.

As the permutation is not important, you can consider ai = i, where the permutation is a1, a2, ..., an.

Output

If it is not possible to swap all pairs of positions so that the permutation stays the same, print "NO",

Otherwise print "YES", then print <image> lines: the i-th of these lines should contain two integers a and b (a < b) — the positions where the i-th swap is performed.

Examples

Input

3


Output

NO


Input

1


Output

YES

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline T parse(const string &s) {
  T x;
  stringstream ss(s);
  ss >> x;
  return x;
}
const double EPS = 1e-12;
const int INF = 1000 * 1000 * 1000;
const long long LINF = INF * 1ll * INF;
const double DINF = 1e200;
const double PI = 3.1415926535897932384626433832795l;
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
long long gcdex(long long a, long long b, long long &x, long long &y) {
  if (!a) {
    x = 0, y = 1;
    return b;
  }
  long long k = b / a;
  long long g = gcdex(b - k * a, a, y, x);
  x -= k * y;
  return g;
}
long long inv(long long a, long long m) {
  long long x, y, g;
  g = gcdex(a, m, x, y);
  return (x % (m / g) + m / g) % m / g;
}
long long crt(long long a1, long long m1, long long a2, long long m2) {
  long long a = (a2 - a1 % m2 + m2) % m2;
  long long x, y, g;
  g = gcdex(m1, m2, x, y);
  if (a % g) return -1;
  long long m = m1 / g * m2;
  x *= a / g;
  x = (x % m2 + m2) % m2;
  long long r = (a1 + x * m1) % m;
  assert(r % m1 == a1 && r % m2 == a2);
  return r;
}
long long powmod(long long a, long long p, long long m) {
  long long r = 1;
  while (p) {
    if (p & 1) r = r * a % m;
    p >>= 1;
    a = a * a % m;
  }
  return r;
}
bool isprime(long long a) {
  if (a <= 1) return false;
  for (long long i = 2; i * i <= a; ++i) {
    if (a % i == 0) return false;
  }
  return true;
}
long long sqrtup(long long a) {
  if (!a) return 0;
  long long x = max(0ll, (long long)sqrt((double)a));
  while (x * x >= a) --x;
  while ((x + 1) * (x + 1) < a) ++x;
  return x + 1;
}
long long isqrt(long long a) {
  long long x = max(0ll, (long long)sqrt((double)a));
  while (x * x > a) --x;
  while ((x + 1) * (x + 1) <= a) ++x;
  return x;
}
long long sgn(long long x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
template <class T>
ostream &operator<<(ostream &s, const vector<T> &v);
template <class A, class B>
ostream &operator<<(ostream &s, const pair<A, B> &p);
template <class K, class V>
ostream &operator<<(ostream &s, const map<K, V> &m);
template <class T, size_t N>
ostream &operator<<(ostream &s, const array<T, N> &a);
template <class... T>
ostream &operator<<(ostream &s, const tuple<T...> &t);
template <class T>
ostream &operator<<(ostream &s, const vector<T> &v) {
  s << '[';
  for (int i = 0; i < (((int)(v).size())); ++i) {
    if (i) s << ',';
    s << v[i];
  }
  s << ']';
  return s;
}
template <class A, class B>
ostream &operator<<(ostream &s, const pair<A, B> &p) {
  s << "(" << p.first << "," << p.second << ")";
  return s;
}
template <class K, class V>
ostream &operator<<(ostream &s, const map<K, V> &m) {
  s << "{";
  bool f = false;
  for (const auto &it : m) {
    if (f) s << ",";
    f = true;
    s << it.first << ": " << it.second;
  }
  s << "}";
  return s;
}
template <class T>
ostream &operator<<(ostream &s, const set<T> &m) {
  s << "{";
  bool f = false;
  for (const auto &it : m) {
    if (f) s << ",";
    f = true;
    s << it;
  }
  s << "}";
  return s;
}
template <class T>
ostream &operator<<(ostream &s, const multiset<T> &m) {
  s << "{";
  bool f = false;
  for (const auto &it : m) {
    if (f) s << ",";
    f = true;
    s << it;
  }
  s << "}";
  return s;
}
template <class T, class V, class C>
ostream &operator<<(ostream &s, const priority_queue<T, V, C> &q) {
  auto a = q;
  s << "{";
  bool f = false;
  while (!a.empty()) {
    if (f) s << ",";
    f = true;
    s << a.top();
    a.pop();
  }
  return s << "}";
}
template <class T, size_t N>
ostream &operator<<(ostream &s, const array<T, N> &a) {
  s << '[';
  for (int i = 0; i < (((int)(a).size())); ++i) {
    if (i) s << ',';
    s << a[i];
  }
  s << ']';
  return s;
}
template <size_t n, class... T>
struct put1 {
  static ostream &put(ostream &s, const tuple<T...> &t) {
    s << get<sizeof...(T) - n>(t);
    if (n > 1) s << ',';
    return put1<n - 1, T...>::put(s, t);
  }
};
template <class... T>
struct put1<0, T...> {
  static ostream &put(ostream &s, const tuple<T...> &t) { return s; }
};
template <class... T>
ostream &operator<<(ostream &s, const tuple<T...> &t) {
  s << "(";
  put1<sizeof...(T), T...>::put(s, t);
  s << ")";
  return s;
}
ostream &put2(ostream &s, const tuple<> &t) { return s; }
template <class U>
ostream &put2(ostream &s, const tuple<U> &t) {
  return s << get<0>(t);
}
template <class U, class V, class... T>
ostream &put2(ostream &s, const tuple<U, V, T...> &t) {
  return s << t;
}
vector<pair<int, int> > solve(int n) {
  vector<pair<int, int> > pre4 = {{0, 1}, {0, 2}, {1, 3},
                                  {0, 3}, {1, 2}, {2, 3}};
  vector<pair<int, int> > pre5 = {{0, 1}, {0, 2}, {0, 3}, {1, 2}, {1, 4},
                                  {2, 4}, {0, 4}, {2, 3}, {1, 3}, {3, 4}};
  vector<pair<int, int> > R = n % 4 ? pre5 : pre4;
  for (int k = 4 + n % 4; k < n; k += 2) {
    if (n % 2) R.push_back(make_pair(0, k));
    R.push_back(make_pair(k, k + 1));
    if (n % 2) R.push_back(make_pair(0, k + 1));
    for (int i = n % 2; i < k; i += 2) {
      R.push_back(make_pair(i, k));
      R.push_back(make_pair(i + 1, k + 1));
      R.push_back(make_pair(i, k + 1));
      R.push_back(make_pair(i + 1, k));
    }
  }
  assert(((int)(R).size()) == n * (n - 1) / 2);
  vector<int> P(n);
  iota((P).begin(), (P).end(), 0);
  for (pair<int, int> s : R) swap(P[s.first], P[s.second]);
  assert(is_sorted((P).begin(), (P).end()));
  {
    vector<pair<int, int> > T = R;
    sort((T).begin(), (T).end());
    int q = 0;
    for (int i = 0; i < (n); ++i) {
      for (int j = i + 1; j < n; ++j) {
        assert(T[q] == make_pair(i, j));
        ++q;
      }
    }
  }
  return R;
}
int main(int argc, char **argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(20);
  int n;
  cin >> n;
  if (n % 4 / 2) {
    cout << "NO";
    return 0;
  }
  if (n == 1) {
    cout << "YES";
    return 0;
  }
  cout << "YES";
  vector<pair<int, int> > R = solve(n);
  for (int i = 0; i < (((int)(R).size())); ++i) {
    cout << '\n' << R[i].first + 1 << ' ' << R[i].second + 1;
  }
  return 0;
}