Codeforces Round #410 (Div. 2) C

Description

Mike has a sequence A = [a₁, a₂, ..., a_n] of length n. He considers the sequence B = [b₁, b₂, ..., b_n] beautiful if the gcd of all its elements is bigger than 1, i.e. .

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 a_i, a_i + 1and put numbers a_i - a_i + 1, a_i+ a_i + 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.

 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, ..., a_n (1 ≤ a_i ≤ 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 .

In the second example the gcd of the sequence is already greater than 1.

题意：要使得，我们可以这样修改 a_i - a_i + 1, a_i+ a_i + 1，问最少修改次数

解法：按照上面操作把所有的数字变成偶数就行了

 #include<bits/stdc++.h>
 using namespace std;
 #define ll long long
 const int maxn=;
 int x[maxn];
 int n;
 int num;
 int sum;
 int main()
 {
     cin>>n;
     cin>>x[];
     num=x[];
     for(int i=; i<=n; i++)
     {
         cin>>x[i];
         num=__gcd(x[i],num);
     }
    // cout<<num<<endl;
     if(num>)
     {
         cout<<"YES\n0\n";
         return ;
     }
     for(int i=; i<n; i++)
     {
         while(x[i]%)
         {
             int pos=x[i];
             x[i]-=x[i+];
             x[i+]+=pos;
             sum++;
         }
     }
    // cout<<x[n]<<endl;
     while(x[n]%)
     {
         int pos=x[n-];
         x[n-]-=x[n];
         x[n]+=pos;
         sum++;
     }
     cout<<"YES\n";
     cout<<sum<<endl;
     return ;
 }