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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {
    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer buffer;

    private static int gcd(int a, int b) {
    	while (b > 0) {
    		int tmp = a;
    		a = b;
    		b = tmp % b;
    	}
    	return a;
    }

    public static void solve() {
        int n=ni();
        int[]a = ni(n);

        int totalGCD = gcd(a[0],a[1]);
        int pos = 2;
        while (pos<n && totalGCD>1) {
            totalGCD = gcd(totalGCD, a[pos]);
            pos++;
        }

        if (totalGCD>1) {
            out.println("YES");
            out.println(0);
            return;
        }

        int moves = 0;
        pos = 0;
        while (pos<n-1) {
            if (a[pos]%2==0) {
                pos++;
                continue;
            }
            if (a[pos+1]%2==1) {
                moves+=1;
                pos+=2;
            } else {
                moves+=2;
                pos+=2;
            }
        }

        if (pos==n-1 && a[n-1]%2==1) moves+=2;
        out.println("YES");
        out.println(moves);
    }


    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    static String next() {
        while (buffer == null || !buffer.hasMoreElements()) {
            try {
                buffer = new StringTokenizer(in.readLine());
            } catch (IOException e) {
                //
            }
        }
        return buffer.nextToken();
    }

    static int ni() {
        return Integer.parseInt(next());
    }

    static long nl() {
        return Long.parseLong(next());
    }

    static double nd() {
        return Double.parseDouble(next());
    }

    static String ns() {
        return next();
    }

    static int[] ni(int n) {
        int[] res = new int[n];
        for (int i = 0; i < n; i++)
            res[i] = Integer.parseInt(next());
        return res;
    }

    static long[] nl(int n) {
        long[] res = new long[n];
        for (int i = 0; i < n; i++)
            res[i] = Long.parseLong(next());
        return res;
    }

    static double[] nd(int n) {
        double[] res = new double[n];
        for (int i = 0; i < n; i++)
            res[i] = Double.parseDouble(next());
        return res;
    }

    static String[] ns(int n) {
        String[] res = new String[n];
        for (int i = 0; i < n; i++)
            res[i] = next();
        return res;
    }


}