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.StringTokenizer;

/**
 * @author Aydar Gizatullin a.k.a. lightning95, aydar.gizatullin@gmail.com
 *         Created on 17.02.17.
 */
public class ProbC {

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

    private void solve() {
        int n = rw.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; ++i)
            a[i] = rw.nextInt();
        int ans = 0;
        int gcd = 0;
        for (int i = 0; i < n; ++i) {
            gcd = gcd(gcd, a[i]);
        }
        if (gcd > 1) {
            rw.println("YES\n0");
            return;
        }
        for (int i = 0; i < n; ) {
            while (Math.abs(a[i] % 2) == 1) {
                if (i + 1 < n) {
                    int x = a[i];
                    a[i] -= a[i + 1];
                    a[i + 1] += x;
                } else {
                    int x = a[i - 1];
                    a[i - 1] -= a[i];
                    a[i] += x;
                    --i;
                }
                ans++;
            }
            ++i;
        }
        for (int i = 0; i < n; ++i)
            System.err.print(a[i] + " ");
        rw.println("YES");
        rw.println(ans);
    }

    private RW rw;
    private String FILE_NAME = "file";

    public static void main(String[] args) {
        new ProbC().run();
    }

    private void run() {
        rw = new RW(FILE_NAME + ".in", FILE_NAME + ".out");
        solve();
        rw.close();
    }

    private class RW {
        private StringTokenizer st;
        private PrintWriter out;
        private BufferedReader br;
        private boolean eof;

        RW(String inputFile, String outputFile) {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));

            File f = new File(inputFile);
            if (f.exists() && f.canRead()) {
                try {
                    br = new BufferedReader(new FileReader(inputFile));
                    out = new PrintWriter(new FileWriter(outputFile));
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }

        private String nextLine() {
            String s = "";
            try {
                s = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return s;
        }

        private String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    eof = true;
                    return "-1";
                }
            }
            return st.nextToken();
        }

        private long nextLong() {
            return Long.parseLong(next());
        }

        private int nextInt() {
            return Integer.parseInt(next());
        }

        private void println() {
            out.println();
        }

        private void println(Object o) {
            out.println(o);
        }

        private void print(Object o) {
            out.print(o);
        }

        private void close() {
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
            out.close();
        }
    }
}