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.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        new Main().run(in, out);
        out.close();
    }

    int gcd(int a, int b) {
        if (a == 0) return b;
        return gcd(b%a, a);
    }

    void run(FastScanner in, PrintWriter out) {


        // if there is already a common factor, the cost will be 0
        // if everything is already even, the cost will be 0
        // otherwise, the min cost will be to make everything even
        // consider two numbdrs
        // if even-even, no-op
        // if odd-odd, can convert to even-even in one op
        //    (2*a+1, 2*b+1) -> (2*a+1-(2*b+1), 2*a+1+2*b+1) = (2*a-2*b, 2*a+2*b+2)
        // if even-odd, can convert to even-even in two ops
        //    (2*a, 2*b+1) -> (2a-2b-1, 2a+2b+1) -> (-4b-2, 4a)
        // if odd-even, also 2 moves


        int n = in.nextInt();
        int[] a = new int[n];

        int gcd = -1;
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
            if (gcd == -1) gcd = a[i];
            else gcd = gcd(gcd, a[i]);
        }

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

        int ret = 0;

        // should convert odd-odd first
        for (int i = 0; i+1<n; i++) {
            int mod1 = a[i]%2;
            int mod2 = a[i+1]%2;
            if (mod1 + mod2 == 2) {
                ret += 1;
                a[i]++; a[i+1]++;
            }
        }

        // convert even-odds
        for (int i = 0; i+1<n; i++) {
            int mod1 = a[i]%2;
            int mod2 = a[i+1]%2;
            if (mod1 + mod2 == 1) {
                ret += 2;
                a[i] -= mod1;
                a[i+1] -= mod2;
            }
        }

        out.println("YES");
        out.println(ret);
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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