Cormen — The Best Friend Of a Man CodeForces 732B

Recently a dog was bought for Polycarp. The dog's name is Cormen. Now Polycarp has a lot of troubles. For example, Cormen likes going for a walk.

Empirically Polycarp learned that the dog needs at least k walks for any two consecutive days in order to feel good. For example, if k = 5 and yesterday Polycarp went for a walk with Cormen 2 times, today he has to go for a walk at least 3 times.

Polycarp analysed all his affairs over the next n days and made a sequence of nintegers a₁, a₂, ..., a_n, where a_i is the number of times Polycarp will walk with the dog on the i-th day while doing all his affairs (for example, he has to go to a shop, throw out the trash, etc.).

Help Polycarp determine the minimum number of walks he needs to do additionaly in the next n days so that Cormen will feel good during all the n days. You can assume that on the day before the first day and on the day after the n-th day Polycarp will go for a walk with Cormen exactly k times.

Write a program that will find the minumum number of additional walks and the appropriate schedule — the sequence of integers b₁, b₂, ..., b_n (b_i ≥ a_i), where b_imeans the total number of walks with the dog on the i-th day.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 500) — the number of days and the minimum number of walks with Cormen for any two consecutive days.

The second line contains integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 500) — the number of walks with Cormen on the i-th day which Polycarp has already planned.

Output

In the first line print the smallest number of additional walks that Polycarp should do during the next n days so that Cormen will feel good during all days.

In the second line print n integers b₁, b₂, ..., b_n, where b_i — the total number of walks on the i-th day according to the found solutions (a_i ≤ b_i for all i from 1 to n). If there are multiple solutions, print any of them.

Examples

Input

3 5
2 0 1

Output

4
2 3 2

Input

3 1
0 0 0

Output

1
0 1 0

Input

4 6
2 4 3 5

Output

0
2 4 3 5

没错，就是遛狗，emmmmm。题目所述就是一个人有一个狗“朋友”，每两天的散步次数总和为K；输入n天已经走了的次数，输出还需要多走几次和每天应该的次数。
其实就是后一天加前一天的次数大于等于K：

#include<iostream>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    int arr[500];
    int sum=0;
    cin>>arr[0];
    for(int i=1;i<a;i++)
    {
        cin>>arr[i];
        if(arr[i]+arr[i-1]<b)
        {
            sum+=b-arr[i]-arr[i-1];
            arr[i]=b-arr[i-1];
        }
    }
    cout<<sum<<endl;
    for(int i=0;i<a;i++)
    {
        if(i!=0)
            cout<<' ';
        cout<<arr[i];
    }
}