CodeForces 732B Cormen — The Best Friend Of a Man

B. Cormen — The Best Friend Of a Man

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 = 5and 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 n integers a1, a2, ..., an, where ai 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 integersb1, b2, ..., bn (bi ≥ ai), where bi means 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, ..., an (0 ≤ ai ≤ 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, ..., bn, where bi — the total number of walks on the i-th day according to the found solutions (ai ≤ bi 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

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;
int n,k;
int a[505];
int b[505];
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    int ans=0;
    for(int i=2;i<=n;i++)
    {
        if(a[i]+a[i-1]>=k) continue;
        else
        {
             ans+=(k-a[i]-a[i-1]);
              a[i]+=(k-a[i]-a[i-1]);
        }


    }
    printf("%d
",ans);
    for(int i=1;i<=n;i++)
    {
        if(i!=n)
            printf("%d ",a[i]);
        else
            printf("%d
",a[i]);
    }
    return 0;

}