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 Main {


    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int n = Integer.parseInt(in.readLine());
        int[] a = new int[n];
        int gcd = 0;
        StringTokenizer st = new StringTokenizer(in.readLine());
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(st.nextToken());
            gcd = gcd(a[i], gcd);
        }
        if (gcd != 1) {
            out.println("YES");
            out.println("0");
            out.close();
            return;
        }
        long res = 0;
        for (int i = 0; i < n-1; i++) {
            int odd = 0;
            if (a[i] % 2 == 1) {
                odd++;
            }
            if (a[i+1] % 2 == 1) {
                odd++;
            }
            if (odd == 0) {
                continue;
            }
            if (odd == 2) {
                res++;
                a[i] = 2;
                a[i+1] = 2;
                i++;
            }
        }
        for (int i = 0; i < n; i++) {
            if (a[i] % 2 == 1) {
                res += 2;
            }
        }
        out.println("YES");
        out.println(res);
        out.close();
    }

    public static int gcd(int p, int q) {
        if (q == 0) return p;
        else return gcd(q, p % q);
    }
}
/*


 */
