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 Copy_2_of_Snippet {

	public static void main(String[] args) {
		FastScanner input = new FastScanner();
		int n = input.nextInt();
		int[] arr = new int[n];
		for (int a = 0; a < n; a++) {
			arr[a] = input.nextInt();
		}
		int gcd = arr[0];
		for(int a = 0; a < n; a++){
			gcd = gcd(gcd, arr[a]);
		}
		
		if(gcd != 1){
			System.out.println("YES");
			System.out.println(0);
		}
		else{
			//fix
			int ops = 0;
			for(int a = 0; a < n-1; a++){
				if(arr[a]%2 == 1 && arr[a+1]%2 == 1){
					arr[a] = 0;
					arr[a+1] = 0;
					ops++;
				}
				if(arr[a]%2 == 1 && arr[a+1]%2 == 0){
					//double
					arr[a] = 0;
					arr[a+1] = 0;
					ops+=2;
				}
			}
			if(arr[n-1]%2 == 1 && arr[n-2]%2 == 0){
				ops+=2;
				arr[n-1] = 0;
				arr[n-2] = 0;
			}
			boolean OK = true;
			for(int a = 0; a < n; a++){
				OK=OK&&(arr[a]%2 == 0);
			}
			if(OK){
				System.out.println("YES");
				System.out.println(ops);
			}
			else{
				System.out.println("NO");
			}
		}
	}

	public static int gcd(int x, int y) {
		return y == 0 ? x : gcd(y, x % y);
	}

	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(Reader in) {
			br = new BufferedReader(in);
		}

		public FastScanner() {
			this(new InputStreamReader(System.in));
		}

		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());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String readNextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		int[] readIntArray(int n) {
			int[] a = new int[n];
			for (int idx = 0; idx < n; idx++) {
				a[idx] = nextInt();
			}
			return a;
		}

		long[] readLongArray(int n) {
			long[] a = new long[n];
			for (int idx = 0; idx < n; idx++) {
				a[idx] = nextLong();
			}
			return a;
		}
	}
}

/*
 * 3 3 1 3 3 1 2 1 2 3 100000 1 3 1000000
 */

/*
 * 3 3 1 2 3 1 2 1 2 3 100000 1 3 1000000
 */

/*
 * 3 3 1 1 3 1 2 1 2 3 100000 1 3 1000000
 */