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.io.*;
import java.util.*;

public class solver2 {

  static BufferedReader in;
  static PrintWriter out;
  static StringTokenizer tok = new StringTokenizer("");

  public static void main(String[] args) throws IOException {
    String filename = "";
    if (filename.isEmpty()) {
      in = new BufferedReader(new InputStreamReader(System.in));
      out = new PrintWriter(System.out);
    } else {
      in = new BufferedReader(new FileReader(filename + ".in"));
      out = new PrintWriter(filename + ".out");
    }
    solve();
    out.close();
  }

  static String readString() throws IOException {
    while (!tok.hasMoreTokens()) {
      tok = new StringTokenizer(in.readLine());
    }
    return tok.nextToken();
  }

  static int readInt() throws IOException {
    return Integer.parseInt(readString());
  }

  static long readLong() throws IOException {
    return Long.parseLong(readString());
  }


  private static void solve() throws IOException {
    int n = readInt();
    int[] a = new int[n];
    int g = 0;
    for (int i = 0; i < n; i++) {
      a[i] = readInt();
      g = gcd(g, a[i]);
    }
    if (g > 1) {
      out.println("YES");
      out.print(0);
      return;
    }
    int ans = 0;
    for (int i = 0; i < n - 1; i++) {
      if (a[i] % 2 == 0) continue;
      if (a[i + 1] % 2 == 0) {
        ans += 2;
        a[i] = 2;
        a[i + 1] = 2;
      } else {
        ans++;
        a[i] = 2;
        a[i + 1] = 2;
      }
      i++;
    }
    if (a[n - 1] % 2 != 0) {
      ans += 2;
    }
    out.println("YES");
    out.print(ans);
  }

  static int gcd(int a, int b) {
    while (a > 0 && b > 0) {
      if (a > b) {
        a %= b;
      } else {
        b %= a;
      }
    }
    return a + b;
  }


}