Problem:
Score : 300 points

Problem Statement

There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N.
Spot i is at the point with coordinate A_i.
It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis.

You planned a trip along the axis.
In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0.

However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i.
You will visit the remaining spots as planned in the order they are numbered.
You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned.

For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.

Constraints

2 \leq N \leq 10^5

-5000 \leq A_i \leq 5000 (1 \leq i \leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print N lines.
In the i-th line, print the total cost of travel during the trip when the visit to Spot i is canceled.

Sample Input 1

3
3 5 -1

Sample Output 1

12
8
10

Spot 1, 2 and 3 are at the points with coordinates 3, 5 and -1, respectively.
For each i, the course of the trip and the total cost of travel when the visit to Spot i is canceled, are as follows:

For i = 1, the course of the trip is 0 \rightarrow 5 \rightarrow -1 \rightarrow 0 and the total cost of travel is 5 + 6 + 1 = 12 yen.

For i = 2, the course of the trip is 0 \rightarrow 3 \rightarrow -1 \rightarrow 0 and the total cost of travel is 3 + 4 + 1 = 8 yen.

For i = 3, the course of the trip is 0 \rightarrow 3 \rightarrow 5 \rightarrow 0 and the total cost of travel is 3 + 2 + 5 = 10 yen.

Sample Input 2

5
1 1 1 2 0

Sample Output 2

4
4
4
2
4

Sample Input 3

6
-679 -2409 -3258 3095 -3291 -4462

Sample Output 3

21630
21630
19932
8924
21630
19288

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	void run() {
		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();

		//始点から出て、始点に戻ってこないといけないのでA[0]とA[N+1]の分、2コ多く確保する。
		int[] A=new int[N+2];

		//始点兼終点は位置0
		A[0]=A[N+1]=0;

		//A[1]～A[N]まで入力を受け取る
		for (int i = 1; i < N+1; i++) {
			A[i]=sc.nextInt();
		}

		//まずは全点の総和をとっておき、各点については計算しておいた総和のとの差を考える
		//ことで計算量を省略する。
		
		int sum=0;
		for (int i = 1; i < N+2; i++) {
			sum+=Math.abs(A[i]-A[i-1]);
		}

		//↑と変わるのは点iの周りだけ
		//具体的にはi-1→i→i+1がなくなり、新しくi-1→i+1へ経路ができる。
		
		for (int i = 1; i < N+1; i++) {

			int ans=sum-Math.abs(A[i-1]-A[i])-Math.abs(A[i]-A[i+1])+Math.abs(A[i-1]-A[i+1]);

			System.out.println(ans);
		}


	}

	//以下テンプレート

	public static int[] extgcd(int a, int b) {

		int x0 = 1;
		int x1 = 0;

		int y0 = 0;
		int y1 = 1;

		while (b != 0) {
			int q = a / b;
			int r = a % b;

			int x2 = x0 - q * x1;
			int y2 = y0 - q * y1;

			a = b;
			b = r;

			x0 = x1;
			x1 = x2;

			y0 = y1;
			y1 = y2;
		}

		return new int[] { a, x0, y0 };

	}

	static int gcd(int a, int b) {

		if (b == 0)
			return a;

		if (a < b) {
			int t = a;
			a = b;
			b = t;
		}

		return gcd(b, a % b);

	}

	static int lcm(int a, int b) {
		return a * b / gcd(a, b);
	}

	static void swap(int[] a) {
		int t;

		t = a[0];
		a[0] = a[1];
		a[1] = t;

		return;
	}

	static <T> void output(List<T> list) {

		for (int i = 0; i < list.size(); i++) {
			System.out.print(list.get(i));

			if (i != list.size() - 1) {
				System.out.print(" ");
			} else {
				nl();
			}
		}

	}

	static void output(String[][] str) {

		for (int i = 0; i < str.length; i++) {
			for (int j = 0; j < str[i].length; j++) {
				print(str[i][j]);
			}

			nl();
		}

	}

	static void output(boolean flg, String Yes, String No) {

		if (flg) {
			pln(Yes);
		} else {
			pln(No);
		}

	}

	static void output(String[][] str, int digit) {

		String dig = "%" + String.valueOf(digit) + "s";

		for (int i = 0; i < str.length; i++) {
			for (int j = 0; j < str[i].length; j++) {
				System.out.printf(dig, str[i][j]);
			}
			nl();
		}

	}

	static void pln(String str) {
		System.out.println(str);
	}

	static void pln(int x) {
		System.out.println(x);
	}

	static void print(String str) {
		System.out.print(str);
	}

	static void print(int x) {
		System.out.print(x);
	}

	static void print(String str, int times) {
		for (int i = 0; i < times; i++) {
			print(str);
		}
	}

	static void print(int x, int times) {
		for (int i = 0; i < times; i++) {
			print(x);
		}
	}

	static void nl() {
		System.out.println();
	}

}