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;
const int M = 100010;
vector<int> primes;
int factors[M];
void gen() {
  for (int i = 2; i < M; i++) {
    if (factors[i] == 0) {
      factors[i] = i;
      primes.push_back(i);
    }
    for (int p : primes) {
      if (i * p >= M) break;
      factors[i * p] = p;
      if (i % p == 0) break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  gen();
  int n;
  cin >> n;
  vector<int> ps;
  int target = n - 1;
  int Tb = n;
  for (int p : primes) {
    if (n % p == 0) ps.push_back(p);
    while (n % p == 0) n /= p;
  }
  if (n > 1) ps.push_back(n);
  if (ps.size() == 1) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    int a = ps[ps.size() - 1];
    int b = ps[ps.size() - 2];
    vector<pair<int, int>> pb;
    for (int i = 0; i * a <= target; i++) {
      int j = (target - i * a) / b;
      if (j * b == target - i * a) {
        if (i > 0) pb.push_back({i, Tb / a});
        if (j > 0) pb.push_back({j, Tb / b});
        break;
      }
    }
    if (pb.size() > 0) {
      cout << pb.size() << endl;
      for (auto& p : pb) {
        cout << p.first << " " << p.second << endl;
      }
    } else {
      assert(false);
    }
  }
}