E. By Elevator or Stairs?

链接：https://codeforces.com/contest/1249/problem/E

You are planning to buy an apartment in a nn-floor building. The floors are numbered from 11 to nn from the bottom to the top. At first for each floor you want to know the minimum total time to reach it from the first (the bottom) floor.

Let:

aiai for all ii from 11 to n−1n−1 be the time required to go from the ii-th floor to the (i+1)(i+1)-th one (and from the (i+1)(i+1)-th to the ii-th as well) using the stairs;
bibi for all ii from 11 to n−1n−1 be the time required to go from the ii-th floor to the (i+1)(i+1)-th one (and from the (i+1)(i+1)-th to the ii-th as well) using the elevator, also there is a value cc — time overhead for elevator usage (you need to wait for it, the elevator doors are too slow!).

In one move, you can go from the floor you are staying at xx to any floor yy (x≠yx≠y) in two different ways:

If you are using the stairs, just sum up the corresponding values of aiai. Formally, it will take ∑i=min(x,y)max(x,y)−1ai∑i=min(x,y)max(x,y)−1ai time units.
If you are using the elevator, just sum up cc and the corresponding values of bibi. Formally, it will take c+∑i=min(x,y)max(x,y)−1bic+∑i=min(x,y)max(x,y)−1bi time units.

You can perform as many moves as you want (possibly zero).

So your task is for each ii to determine the minimum total time it takes to reach the ii-th floor from the 11-st (bottom) floor.

Input

The first line of the input contains two integers nn and cc (2≤n≤2⋅105,1≤c≤10002≤n≤2⋅105,1≤c≤1000) — the number of floors in the building and the time overhead for the elevator rides.

The second line of the input contains n−1n−1 integers a1,a2,…,an−1a1,a2,…,an−1 (1≤ai≤10001≤ai≤1000), where aiai is the time required to go from the ii-th floor to the (i+1)(i+1)-th one (and from the (i+1)(i+1)-th to the ii-th as well) using the stairs.

The third line of the input contains n−1n−1 integers b1,b2,…,bn−1b1,b2,…,bn−1 (1≤bi≤10001≤bi≤1000), where bibi is the time required to go from the ii-th floor to the (i+1)(i+1)-th one (and from the (i+1)(i+1)-th to the ii-th as well) using the elevator.

Output

Print nn integers t1,t2,…,tnt1,t2,…,tn, where titi is the minimum total time to reach the ii-th floor from the first floor if you can perform as many moves as you want.

Examples

input

Copy

10 2
7 6 18 6 16 18 1 17 17
6 9 3 10 9 1 10 1 5

output

Copy

0 7 13 18 24 35 36 37 40 45

input

Copy

10 1
3 2 3 1 3 3 1 4 1
1 2 3 4 4 1 2 1 3

output

Copy

0 2 4 7 8 11 13 14 16 17

题解：

一道dp题，只要注意有先坐电梯再走楼梯再坐电梯的操作（虽然俺也不知道为什么有这样神奇的操作，但的确有这样的最优解）

代码：

#include<bits/stdc++.h>
using namespace std;
long long n,m,s;
long long a[1000001],b[1000001],dp[1000001][2];
int main()
{
	cin>>n>>m;
	s=0;
	for(int i=1;i<n;i++)
	{
		cin>>a[i];
	}
	for(int i=1;i<n;i++)
	{
		cin>>b[i];
	}
	cout<<0;
	dp[0][0]=0;
	dp[0][1]=m;
	for(int i=1;i<n;i++)
	{
		dp[i][0]=min(dp[i-1][0]+a[i],dp[i-1][1]+b[i]);
		dp[i][1]=min(dp[i-1][0]+a[i]+m,dp[i-1][1]+b[i]);
		cout<<" "<<min(dp[i][0],dp[i][1]);
	}
}