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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.NoSuchElementException;

public class C {
	int N;
	int[] a;

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

	public void solve() {
		N = nextInt();
		a = new int[N];
		for(int i = 0;i < N;i++){
			a[i] = nextInt();
		}

		int ans = 0;
		int g = a[0];

		for(int i = 1;i < N;i++){
			g = gcd(g,a[i]);
		}

		if(g > 1){
			out.println("YES");
			out.println(ans);
			return;
		}

		for(int i = 0;i < N-1;i++){
			while(a[i] % 2 == 1){
				int pre = Math.abs(a[i] - a[i+1]);
				int next = a[i] + a[i+1];
				a[i] = pre;
				a[i+1] = next;
				ans++;
			}
		}

		if(a[N-1] % 2 == 1)ans += 2;

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

	public static void main(String[] args) {
		out.flush();
		new C().solve();
		out.close();
	}

	/* Input */
	private static final InputStream in = System.in;
	private static final PrintWriter out = new PrintWriter(System.out);
	private final byte[] buffer = new byte[2048];
	private int p = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (p < buflen)
			return true;
		p = 0;
		try {
			buflen = in.read(buffer);
		} catch (IOException e) {
			e.printStackTrace();
		}
		if (buflen <= 0)
			return false;
		return true;
	}

	public boolean hasNext() {
		while (hasNextByte() && !isPrint(buffer[p])) {
			p++;
		}
		return hasNextByte();
	}

	private boolean isPrint(int ch) {
		if (ch >= '!' && ch <= '~')
			return true;
		return false;
	}

	private int nextByte() {
		if (!hasNextByte())
			return -1;
		return buffer[p++];
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = -1;
		while (isPrint((b = nextByte()))) {
			sb.appendCodePoint(b);
		}
		return sb.toString();
	}

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

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

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