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. .

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.

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 .

In the second example the gcd of the sequence is already greater than 1.
題意：給定一組序列，通過若干次變化是指成為一個美麗的序列。美麗序列的定義是，數(shù)組中所有數(shù)字的最大公因數(shù)大于1。問是否能轉(zhuǎn)換成這樣的一個序列，并且最少的變換次數(shù)是多少。
思路：一定可以轉(zhuǎn)換成。
如果這個序列一開始就是最大公因數(shù)大于1，那么就不需要變換。如果不是的話，就有這樣的幾種情況。奇奇，偶偶，奇偶，偶奇。對于偶偶來說，不用管。對于其他三種情況來說
①奇奇：轉(zhuǎn)換成偶偶，只需要一步。
②奇偶，偶奇：轉(zhuǎn)換成偶偶需要兩步。
因為求最少的步數(shù)，所以我們應該應該多轉(zhuǎn)換奇奇，其次是奇偶，偶奇。
所以我們先找數(shù)組中的奇奇，然后再找奇偶，或者偶奇。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=1e5+100; int a[maxx]; int n;int main() {scanf("%d",&n);scanf("%d",&a[1]);int cnt=a[1];for(int i=2;i<=n;i++) {scanf("%d",&a[i]);cnt=__gcd(cnt,a[i]);}ll sum=0;if(cnt>1) {cout<<"YES"<<endl<<sum<<endl;return 0;}for(int i=1;i<n;i++){if((a[i]&1)&&(a[i+1]&1)) sum++,a[i]+=1,a[i+1]+=1;}for(int i=1;i<n;i++){if((a[i]+a[i+1])&1) sum+=2,a[i]=a[i+1]=2;//都轉(zhuǎn)換成偶數(shù)。。}cout<<"YES"<<endl<<sum<<endl; }

努力加油a啊，(^o)/~

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