【56.74%】【codeforces 732B】Cormen --- The Best Friend Of a Man

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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 = 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 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 integers b1, 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

【题解】



只要前一天和今天的次数和没有大于等于k，则增加当前天的次数，直至前一天和今天的次数和等于k。

勇敢贪吧。

#include <cstdio>

int a[600];
int n, k;

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    a[0] = k; a[n + 1] = k;
    int add = 0;
    for (int i = 1; i <= n; i++)
    {
        int num = a[i - 1] + a[i];
        if (num < k)
        {
            add += (k - num);
            a[i] += (k - num);
        }
    }
    printf("%d\n", add);
    for (int i = 1; i <= n; i++)
        printf("%d%c", a[i], i == n ? '\n' : ' ');
    return 0;
}