Codeforces Round #376 (Div. 2) A && B_grigoriy, like the hero of one famous comedy film-优快云博客

Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he receivedembosser and was to take stock of the whole exposition.

Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:

$\text{[math]}$

After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.

Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.

Input

The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.

Output

Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.

Examples

input
zeus

output
18

input
map

output
35

input
ares

output
34

Note

$\text{[math]}$

To print the string from the first sample it would be optimal to perform the following sequence of rotations:

from 'a' to 'z' (1 rotation counterclockwise),
from 'z' to 'e' (5 clockwise rotations),
from 'e' to 'u' (10 rotations counterclockwise),
from 'u' to 's' (2 counterclockwise rotations).

In total, 1 + 5 + 10 + 2 = 18 rotations are required.

题意：直接看hint就知道题意了；

思路：很水的一题，要么直接减，要么就是小的+26减大的取一个最小值，把输出字符串首字母改成a就是了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int main()
{
    char str[105];
    scanf("%s", str+1);
    str[0] = 'a';
    int ans = 0;
  //  cout << str << endl;
    for(int i = 1; i < strlen(str); i++)
    {
     //   int flag = 0;
      //  if(str[i-1] > str[i]) swap(str[i-1], str[i]), flag = 1;
        ans += min(min(abs(str[i-1]-str[i]), abs(str[i-1]+26-str[i])), abs((str[i]+26-str[i-1])));
        //if(flag)  swap(str[i-1], str[i]);
    }


        cout << ans << endl;
    return 0;
}

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：要么给一个元素直接-2，要么给相邻两个元素同时-1；

思路：还是比较水的，只要某个元素%2 == 1他就一定要-1，所以从前遍历，找到一个 == 1的就把他的后面-1，如果这时候他后面<0了，就不能全部清空，根本不用管%2 == 0的，因为没问你最少的次数。。可以随便拿的，就是不要忘记最后一个元素要特判下，，，

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 2e5 + 5;
int a[maxn], n;
int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
    }
    for(int i = 2; i <= n; i++)
    {
        if(a[i-1] % 2) a[i]--;
        if(a[i] < 0) {printf("NO\n"); return 0;}
    }
    if(a[n] % 2) printf("NO\n");  //因为只是看的前面那个元素，如果最后一个元素%2 == 1，没人跟他一起-1，所以no
    else printf("YES\n");
    return 0;
}