Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful if the gcd of all its elements is bigger than 1, i.e. <image>. 

Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers ai, ai + 1 and put numbers ai - ai + 1, ai + ai + 1 in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.

<image> is the biggest non-negative number d such that d divides bi for every i (1 ≤ i ≤ n).

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.

Output

Output on the first line "YES" (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and "NO" (without quotes) otherwise.

If the answer was "YES", output the minimal number of moves needed to make sequence A beautiful.

Examples

Input

2
1 1


Output

YES
1


Input

3
6 2 4


Output

YES
0


Input

2
1 3


Output

YES
1

Note

In the first example you can simply make one move to obtain sequence [0, 2] with <image>.

In the second example the gcd of the sequence is already greater than 1. 
import java.util.*;
import java.util.stream.*;
import java.io.*;
import java.math.*;
import java.awt.geom.*;
public class Main {
  public static final BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
  public static final PrintWriter outWriter = new PrintWriter(System.out);
  public static long gcd(long[] a) {
    long out = a[0];
    for (int i = 1; i < a.length; i++) {
      out = gcd(out, a[i]);
    }
    return out;
  }
  public static long gcd(long a, long b) { return b < 0 ? gcd(a,-b) : a < 0 ? gcd(-a,b) : b==0 ? a : gcd(b, a%b); }
  public static void main() {
    int n = nextInt();
    long[] a = new long[n];
    for (int i = 0; i < a.length; i++) { a[i] = nextLong(); }
    outWriter.println("YES");
    if (gcd(a) > 1) {
      outWriter.println(0);
    } else {
      int count = 0;
      for (int i = 1; i < a.length; i++) {
        if (a[i] % 2 == 1 && a[i-1]%2 == 1) {
          count++;
          a[i] = a[i-1] = 2;
        }
      }
      for (int i = 0; i < a.length; i++) {
        if (a[i] % 2 == 1) {
          count+=2;
        }
      }
      outWriter.println(count);
    }
  }
  public static void main(String[] args) { main(); outWriter.flush(); }
  public static StringTokenizer tokenizer = null;
  public static String nextLine() {
    try { return buffer.readLine(); } catch (IOException e) { throw new UncheckedIOException(e); }
  }
  public static String next() {
    while (tokenizer == null || !tokenizer.hasMoreElements()) { tokenizer = new StringTokenizer(nextLine()); }
    return tokenizer.nextToken();
  }
  public static int nextInt() { return Integer.parseInt(next()); }
  public static long nextLong() { return Long.parseLong(next()); }
  public static double nextDouble() { return Double.parseDouble(next()); }
}
