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.*;
import java.awt.geom.*;
import java.math.*;

public class C {

    static final Scanner in = new Scanner(System.in);
    static final PrintWriter out = new PrintWriter(System.out,false);
    static boolean debug = false;

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

    static void solve() {
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i=0; i<n; i++) {
            a[i] = in.nextInt();
        }

        int gcd = a[0];
        for (int i=1; i<n; i++) {
            gcd = gcd(gcd, a[i]);
        }
        if (gcd > 1) {
            out.println("YES");
            out.println(0);
            return;
        }

        long ans = 0;
        for (int i=0; i<n-1; i++) {
            if (a[i]%2 == 1 && a[i+1]%2 == 1) {
                ans += 1;
                a[i] = 2;
                a[i+1] = 2;
            }
        }

        for (int i=0; i<n-1; i++) {
            if (a[i]%2 != a[i+1]%2) {
                ans += 2;
                a[i] = 2;
                a[i+1] = 2;
            }
        }
        out.println("YES");
        // dump(a);
        out.println(ans);
    }

    public static void main(String[] args) {
        debug = args.length > 0;
        long start = System.nanoTime();

        solve();
        out.flush();

        long end = System.nanoTime();
        in.close();
        out.close();
    }

    static void dump(Object... o) { if (debug) System.err.println(Arrays.deepToString(o)); }
}