Problem: You are given a positive integer n.

Find a sequence of fractions (a_i)/(b_i), i = 1 … k (where a_i and b_i are positive integers) for some k such that:

$$$ \begin{cases} $b_i$ divides $n$, $1 < b_i < n$ for $i = 1 … k$ \\\ $1 ≤ a_i < b_i$ for $i = 1 … k$ \\\ \text{$∑_{i=1}^k (a_i)/(b_i) = 1 - 1/n$} \end{cases} $$$

Input

The input consists of a single integer n (2 ≤ n ≤ 10^9).

Output

In the first line print "YES" if there exists such a sequence of fractions or "NO" otherwise.

If there exists such a sequence, next lines should contain a description of the sequence in the following format.

The second line should contain integer k (1 ≤ k ≤ 100 000) — the number of elements in the sequence. It is guaranteed that if such a sequence exists, then there exists a sequence of length at most 100 000.

Next k lines should contain fractions of the sequence with two integers a_i and b_i on each line.

Examples

Input


2


Output


NO


Input


6


Output


YES
2
1 2
1 3

Note

In the second example there is a sequence 1/2, 1/3 such that 1/2 + 1/3 = 1 - 1/6.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class U, class T>
void Max(U &first, T second) {
  if (first < second) first = second;
}
template <class U, class T>
void Min(U &first, T second) {
  if (first > second) first = second;
}
template <class T>
void add(int &a, T b) {
  a = (a + b) % 1000000007;
}
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % 1000000007;
    a = 1LL * a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int pow(int a, int b, int c) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % c;
    a = 1LL * a * a % c;
    b >>= 1;
  }
  return ans;
}
int q[100];
int main() {
  int ca = 0, T, k, i, j, m = 0, K, n;
  scanf("%d", &n);
  m = n;
  int sz = 0;
  for (int i = 2; i < m + 1; i++) {
    if (1LL * i * i > m) break;
    if (m % i) continue;
    while (m % i == 0) {
      m /= i;
    }
    q[sz++] = i;
  }
  if (m > 1) q[sz++] = m;
  if (sz == 1) {
    puts("NO");
    return 0;
  }
  for (int i = 0; i < sz - 1; i++) {
    for (int j = i + 1; j < sz; j++) {
      int first = q[i], second = q[j];
      m = n - 1;
      int a = m / second;
      m %= second;
      while (m % first && a > 1) {
        m += second, a--;
      }
      if (a < n / second && m % first == 0 && m / first < n / first) {
        printf("YES\n2\n");
        printf("%d %d\n%d %d\n", m / first, n / first, a, n / second);
        return 0;
      }
    }
  }
  puts("NO");
}