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 File {
	
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner() {
			br = new BufferedReader(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());
		}
	}
	
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		
	
		int n = sc.nextInt();
		long[] A = new long[n];
		
		for (int i = 0; i < n; i++) {
			A[i] = sc.nextLong();
		}
		
		long gcd = A[0];
		for (int i = 1; i < n; i++) {
			gcd = getGCD(gcd, A[i]);
		}
		
		// GCD is already > 1
		if (gcd > 1) {
			out.println("YES");
			out.println(0);
			out.close();
			return;
		}
		// Only two elements, check if possible after 1 move.
		if (n == 2) {
			int moves = 0;
			
			// Both odd.
			if (A[0] % 2 == 1 && A[1] % 2 == 1) {
				moves = 1; // Both will become even after 1 move.
			}			
			// A[0] is odd, A[1] is even.
			else {
				// Always works in 2 moves at most, when both values are even.
				long next1 = A[0] - A[1];
				long next2 = A[0] + A[1];
				
				if (getGCD(next1, next2) > 1) {
					moves = 1;
				}
				else {
					moves = 2;
				}
			}
			
			out.println("YES");
			out.println(moves);
			out.close();
			return;
		}
		
		// Change all GCD to be 2, and we're guaranteed all elements GCD > 1.
		// Make sure first 2 numbers are even.
		int moves = 0;
		
		// Change all pairs of odds to evens in 1 move.
		for (int i = 0; i + 1 < n; i++) {
			if (A[i] % 2 == 1 && A[i+1] % 2 == 1) {
				moves++;
				A[i] = 0;
				A[i+1] = 0;
			}
		}
		
		// Both are odd.
		if (A[0] % 2 == 1 && A[1] % 2 == 1) {
			moves++;
		}
		// Exactly one is odd.
		else if ((A[0] % 2) != (A[1] % 2)) {
			// Two moves to make them both even.
			// First move, both become odd. Second move, both become even.
			moves += 2; 
		}
		
		for (int i = 2; i < n ; i++) {
			// Only need to make move to make both even.
			if (A[i] % 2 == 1) {
				moves += 2;
			}
		}
		
		out.println("YES");
		out.println(moves);
		
		out.close();
	}
	
	public static long getGCD(long a, long b) {
		if (b == 0) {
			return (long)Math.abs(a);
		}
		if (a == 0) {
			return (long)Math.abs(b);
		}
		
		return getGCD(b, a % b);
	}
}