codeforces 732B Cormen — The Best Friend Of a Man

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B. Cormen — The Best Friend Of a Man

time limit per test

memory limit per test

input

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.

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

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.).

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

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

Input

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

a1, a2, ..., an (0 ≤ ai) — the number of walks with Cormen on the i-th day which Polycarp has already planned.

Output

n

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

题意：最近的两个的和要大于或者等于k，不够的补充，输出补充的数的和，和改变之后的数。参考了别人的想法。

#include<stdio.h>
int main()
{
    int n,k,s[510];
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&s[i]);
    }
    int cnt=0;
    for(int i=1;i<n;i++)
    {
        if(s[i]+s[i-1]<k)
        {
            cnt+=k-(s[i]+s[i-1]);
            s[i]=k-s[i-1];
        }
    }
    printf("%d\n",cnt);
    for(int i=0;i<n;i++)
    {
        if(i==n-1)
            printf("%d\n",s[i]);
        else
            printf("%d ",s[i]);
    }
    return 0;
}