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. 
// <editor-fold defaultstate="collapsed" desc="Jose Nolberto Isac Gonzalez">
//============================================================================
// Name        : Solucion.java
// Author      : Jose Nolberto Isac Gonzalez
// Version     :
// Copyright   : Your copyright notice
// Description : DESOFT Division GRANMA
//============================================================================
// </editor-fold>

import java.util.*;

public class Main {

    static int n, cont;
    static long arr[];
    static int dp[];
    static long aux;

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

    static void solve() {
        arr[n + 1] = 2;
        aux = gcd(arr[1], arr[2]);
        for (int i = 3; i <= n; i++) {
            aux = gcd(aux, arr[i]);
        }
        if (aux > 1) {
            System.out.println("YES");
            System.out.println("0");
            return;
        }
        cont = 0;
        for (int i = 1; i <= n; i++) {
            if (arr[i] % 2 == 0) {
                continue;
            }
            if ((arr[i] % 2 == 1 && arr[i + 1] % 2 == 0) || (arr[i] % 2 == 0 && arr[i + 1] % 2 == 1)) {
                cont += 2;
                arr[i + 1] = 2;
            } else if (arr[i] % 2 == 1 && arr[i + 1] % 2 == 1) {
                cont += 1;
                arr[i + 1] = 2;
            }
        }
        System.out.println("YES");
        System.out.println(cont);
    }

    public static void main(String[] args) {
        Scanner l = new Scanner(System.in);
        n = l.nextInt();
        arr = new long[105000];
        dp = new int[105000];
        for (int i = 1; i <= n; i++) {
            arr[i] = l.nextLong();
        }
        solve();
    }
}
