Codeforces Round #376 (Div. 2) B. Coupons and Discounts

B. Coupons and Discounts

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a number of year and he knows that to get ready for the training session it's not enough to prepare only problems and editorial. As the training sessions lasts for several hours, teams become hungry. Thus, Sereja orders a number of pizzas so they can eat right after the end of the competition.

Teams plan to train for n times during n consecutive days. During the training session Sereja orders exactly one pizza for each team that is present this day. He already knows that there will be ai teams on the i-th day.

There are two types of discounts in Sereja's favourite pizzeria. The first discount works if one buys two pizzas at one day, while the second is a coupon that allows to buy one pizza during two consecutive days (two pizzas in total).

As Sereja orders really a lot of pizza at this place, he is the golden client and can use the unlimited number of discounts and coupons of any type at any days.

Sereja wants to order exactly ai pizzas on the i-th day while using only discounts and coupons. Note, that he will never buy more pizzas than he need for this particular day. Help him determine, whether he can buy the proper amount of pizzas each day if he is allowed to use only coupons and discounts. Note, that it's also prohibited to have any active coupons after the end of the day n.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 200 000) — the number of training sessions.

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 10 000) — the number of teams that will be present on each of the days.

Output

If there is a way to order pizzas using only coupons and discounts and do not buy any extra pizzas on any of the days, then print "YES" (without quotes) in the only line of output. Otherwise, print "NO" (without quotes).

Examples

input

4
1 2 1 2

output

YES

input

3
1 0 1

output

NO

Note

In the first sample, Sereja can use one coupon to buy one pizza on the first and the second days, one coupon to buy pizza on the second and the third days and one discount to buy pizzas on the fourth days. This is the only way to order pizzas for this sample.

In the second sample, Sereja can't use neither the coupon nor the discount without ordering an extra pizza. Note, that it's possible that there will be no teams attending the training sessions on some days.

题意：就是他只有两种买披萨的方式，一种一天买两个，一种连续两天每天买一个，一天可以用多种方式，问可不可以用这两种方式买到所需要的披萨。

思路：首先当我们遇到奇数是我们肯定要用第二种方式，然后让其减一成为偶数，偶数就可以用第第一种方式。到最后全为偶数就证明可以买到所需的披萨。我本来想的是先用第二种买买到不能买，然后再看能不能被第二种买，如果不能就输出NO，所以用我的程序运行3 2 3结果就是NO，但是正确答案是YES，晚上想了好久，WA了好几遍，早上来了换了一种思路一遍就过了，所以当一种思路要是老是错的时候就可以换种思路想一想。

代码：

#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<stdlib.h>
#include<stack>
#include<vector>
#include<string.h>
#include<map>
#define INF 0x3f3f3f3f3f
using namespace std;
int a[200002];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=0;i<n-1;i++)
        {
            if(a[i]%2==1)
            {
                a[i]-=1;
                a[i+1]-=1;
            }
        }
        int flag=1;
        for(int i=0;i<n;i++)
        {
            a[i]%=2;
            if(a[i]!=0)
            {
                flag=0;
            }
        }
        if(flag==0)
        {
            printf("NO\n");
        }
        else
        {
            printf("YES\n");
        }
    }
}