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 integers b1, b2, ..., bn (bi ≥ ai), where bi means the total number of walks with the dog on the i-th day.

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.

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.

3 5
2 0 1

3 1
0 0 0

4 6
2 4 3 5

4
2 3 2

1
0 1 0

0
2 4 3 5

题意：

给出一串数，要求前后连续的两个数的和不小于k，问每个数在符合要求的情况下，最少需要加多少，输出最后的串的值。

官方题解：

If we don‘t make enough walks during days i and i + 1, it‘s better to make an additional walk on day i + 1 because it also counts as a walk during days i + 1 and i + 2 (and if we walk one more time on day i, it won‘t help us in the future). So we can start iterating from the second day (1"=indexed). We will add max(0, k - ai - ai - 1) walks to the day i (and to our answer), so Cormen has enough walks during days i and i - 1. After we have iterated through all days, we can print the answer.

Time complexity: O(n).

#include<bits/stdc++.h>using namespace std;const int maxn = 505;int main(){	int n,k,sum = 0,a[maxn];	scanf("%d%d",&n,&k);	for (int i = 0;i < n;i++)	scanf("%d",&a[i]);	for (int i = 1;i < n;i++)	sum += max(0,k-a[i-1]-a[i]),a[i] +=max(0,k-a[i-1]-a[i]);	printf("%d\n",sum);	printf("%d",a[0]);	for (int i = 1;i < n;i++)	printf(" %d",a[i]);	printf("\n");	return 0;}