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. 
/*
 *created by Kraken on 29-04-2020 at 14:27
 */
//package com.kraken.cf.practice;

import java.util.*;
import java.io.*;

public class C798 {

  public static void main(String[] args) {
    FastReader sc = new FastReader();
    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 = gcd(gcd, a[i]);
    if (gcd > 1) {
      System.out.println("YES\n0");
      return;
    }
    long res = 0;
    int odd = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] % 2 != 0) odd++;
      else {
        res += odd % 2 == 0 ? odd / 2 : odd / 2 + 2;
        odd = 0;
      }
    }
    if (odd > 0) res += odd % 2 == 0 ? odd / 2 : odd / 2 + 2;
    System.out.println("YES\n" + res);
  }

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

  static class FastReader {

    BufferedReader br;

    StringTokenizer st;

    public FastReader() {
      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());
    }

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

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