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.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
/*

2
3 5

2
6 5
1 11
-10 12

3
3 6 2


1   8   5   3
-7  9   5   3
-7  4   14  3


1   8   5   3
1   3   13  3
-2  4   13  3
   

3
1 3 6

3 3 6

5 12
-7 19


3 5 7 11 13
-2 8


5 9


5 9


3 6 7
-3 9 7

6 3 7


*/

public class c {
	static int n;
	static int inf = (int) 1e8;
	static int[] arr;
	public static void main(String[] args) throws IOException {
		FastScanner in = new FastScanner(System.in);
		n = in.nextInt();
		arr = new int[n];
		for(int i=0;i<n;i++)
			arr[i] = in.nextInt();
		System.out.println("YES");
		int tmp = 0;
		for(int i=0;i<n;i++)
			tmp = gcd(tmp, arr[i]);
		if(tmp > 1) {
			System.out.println(0);
			return;
		}
		memo = new int[2][n+1];
		for(int[] arr : memo)
			Arrays.fill(arr, -1);
		System.out.println(solve(0, arr[0]%2));
	}
	static int[][] memo;
	static int solve(int idx, int par) {
		if(idx == n-1) {
			if(par == 0)
				return 0;
			return inf;
		}
		if(memo[par][idx] != -1)
			return memo[par][idx];
		int a = par, b = arr[idx+1]%2;
		int ans = inf;
		for(int amt=0;amt<=2;amt++) {
			if(a % 2 == 0) {
				ans = Math.min(ans,solve(idx+1, b)+amt);
			}
			int diff = (a-b+4)%2;
			int sum = (a+b+4)%2;
			a = diff;
			b = sum;
		}
		return memo[par][idx] = ans;
	}
	static int gcd(int a, int b) {
		return b == 0 ? a : gcd(b,a%b);
	}
	static class FastScanner {
	    BufferedReader br;
	    StringTokenizer st;
	    public FastScanner(InputStream i) {
	        br = new BufferedReader(new InputStreamReader(i));
	    }
	    public String next() throws IOException {
	        while(st == null || !st.hasMoreTokens())
	            st = new StringTokenizer(br.readLine());
	        return st.nextToken();
	    }
	    public int nextInt() throws IOException {
	        return Integer.parseInt(next());
	    }
	    public long nextLong() throws IOException {
	        return Long.parseLong(next());
	    }
	    public double nextDouble() throws IOException {
	        return Double.parseDouble(next());
	    }
	}
}
