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;
const double pi = acos(0.0) * 2.0;
const long double eps = 1e-10;
const int step[8][2] = {{-1, 0}, {0, 1}, {1, 0},  {0, -1},
                        {-1, 1}, {1, 1}, {1, -1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <typename t, typename t1>
t min1(t a, t1 b) {
  return a < b ? a : b;
}
template <typename t, typename... arg>
t min1(t a, arg... arr) {
  return min1(a, min1(arr...));
}
template <typename t, typename t1>
t max1(t a, t1 b) {
  return a > b ? a : b;
}
template <typename t, typename... arg>
t max1(t a, arg... arr) {
  return max1(a, max1(arr...));
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / max(abs1(a), abs1(b)) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  if (na == 0) return 0;
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
template <class T>
inline T lowb(T num) {
  return num & (-num);
}
inline int bitnum(unsigned int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(unsigned long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitnum(long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitmaxl(unsigned int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(unsigned long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
inline int bitmaxl(long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
long long pow(long long n, long long m, long long mod = 0) {
  if (m < 0) return 0;
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
template <class t1, class t2>
inline void add(t1 &a, t2 b, int mod = -1) {
  if (mod == -1) mod = 1000000007;
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
}
template <class t>
void output1(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) cerr << arr[i] << ' ';
  cerr << endl;
}
template <class t>
void output2(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) output1(arr[i]);
}
const int maxn = 1010;
int n;
pair<int, int> nxt[maxn][maxn];
pair<int, int> bef[maxn][maxn];
pair<int, int> tail = make_pair(-1, -1);
void addToTail(int i, int j) {
  if (i > j) swap(i, j);
  if (tail.first != -1) nxt[tail.first][tail.second] = make_pair(i, j);
  bef[i][j] = tail;
  tail = make_pair(i, j);
}
void add4(int n) {
  addToTail(n, n + 1);
  addToTail(n, n + 2);
  addToTail(n + 1, n + 3);
  addToTail(n, n + 3);
  addToTail(n + 1, n + 2);
  addToTail(n + 2, n + 3);
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d", &n);
  if ((n / 2) % 2) {
    puts("NO");
    return 0;
  }
  memset(nxt, -1, sizeof(nxt));
  memset(bef, -1, sizeof(bef));
  for (int i = 0; i < n - 1; i += 4) {
    for (int j = 0; j < i; j++) addToTail(i, j);
    addToTail(i, i + 1);
    for (int j = i - 1; j >= 0; j--) addToTail(i + 1, j);
    addToTail(i, i + 2);
    addToTail(i + 1, i + 3);
    addToTail(i, i + 3);
    addToTail(i + 1, i + 2);
    for (int j = 0; j < i; j++) addToTail(i + 2, j);
    addToTail(i + 2, i + 3);
    for (int j = i - 1; j >= 0; j--) addToTail(i + 3, j);
  }
  if (n % 2) {
    for (int j = 0; j < n / 2 * 2; j += 2) {
      int i = n - 1;
      pair<int, int> rbef = bef[j][j + 1];
      bef[j][i] = rbef;
      nxt[j][i] = make_pair(j, j + 1);
      bef[j][j + 1] = make_pair(j, i);
      if (rbef.first != -1) nxt[rbef.first][rbef.second] = make_pair(j, i);
      pair<int, int> rnxt = nxt[j][j + 1];
      bef[j + 1][i] = make_pair(j, j + 1);
      nxt[j + 1][i] = rnxt;
      nxt[j][j + 1] = make_pair(j + 1, i);
      if (rnxt.first != -1) bef[rnxt.first][rnxt.second] = make_pair(j + 1, i);
    }
  }
  puts("YES");
  pair<int, int> head = make_pair(0, 1);
  while (bef[head.first][head.second].first != -1) {
    head = bef[head.first][head.second];
  }
  if (n == 1) return 0;
  while (head.first != -1) {
    printf("%d %d\n", head.first + 1, head.second + 1);
    head = nxt[head.first][head.second];
  }
  return 0;
}