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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

// 内部定数
#define D_ARRAY_MAX		100000									// 最大配列数

// 内部変数
static int si1Array[D_ARRAY_MAX + 5];							// 配列
static int siACnt;												// 配列数
static int siSum;												// 取りやめない時の合計値

// 内部変数 - テスト用
#ifdef D_TEST
	static int siRes;
	static FILE *szpFpT, *szpFpA;
#endif

// 実行メイン
int
fMain(
	int piTNo					// <I> テスト番号 1～
)
{
	int i;
	char lc1Buf[1024], lc1Out[1024];

	// テストファイルオープン
#ifdef D_TEST
	sprintf(lc1Buf, ".\\Test\\T%d.txt", piTNo);
	szpFpT = fopen(lc1Buf, "r");
	sprintf(lc1Buf, ".\\Test\\A%d.txt", piTNo);
	szpFpA = fopen(lc1Buf, "r");
	siRes = 0;
#endif

	// 観光スポット数取得
	int liCnt;
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpT);
#else
	fgets(lc1Buf, sizeof(lc1Buf), stdin);
#endif
	sscanf(lc1Buf, "%d", &liCnt);

	// 配列取得
	si1Array[0] = 0;
	si1Array[liCnt + 1] = 0;
	for (i = 1; i <= liCnt; i++) {
#ifdef D_TEST
		fscanf(szpFpT, "%d", &si1Array[i]);
#else
		fscanf(stdin, "%d", &si1Array[i]);
#endif
	}
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpT);
#else
	fgets(lc1Buf, sizeof(lc1Buf), stdin);
#endif

	// 配列数
	siACnt = liCnt + 2;

	// 取りやめない時の合計値
	siSum = 0;
	for (i = 0; i < siACnt - 1; i++) {
		siSum += abs(si1Array[i] - si1Array[i + 1]);
	}

	// 取りやめる場合を出力
	for (i = 1; i < siACnt - 1; i++) {

		// 前後の金額減額
		int liSum = siSum;
		liSum -= abs(si1Array[i] - si1Array[i - 1]);
		liSum -= abs(si1Array[i] - si1Array[i + 1]);

		// ２マス移動額
		liSum += abs(si1Array[i - 1] - si1Array[i + 1]);

		// 結果セット
		sprintf(lc1Out, "%d\n", liSum);

		// 結果表示
#ifdef D_TEST
		fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
		if (strcmp(lc1Buf, lc1Out)) {
			siRes = -1;
		}
#else
		printf("%s", lc1Out);
#endif
	}

	// テストファイルクローズ
#ifdef D_TEST
	fclose(szpFpT);
	fclose(szpFpA);
#endif

	// テスト結果
#ifdef D_TEST
	if (siRes == 0) {
		printf("OK %d\n", piTNo);
	}
	else {
		printf("NG %d\n", piTNo);
	}
#endif

	return 0;
}

int
main()
{

#ifdef D_TEST
	int i;
	for (i = D_TEST_SNO; i <= D_TEST_ENO; i++) {
		fMain(i);
	}
#else
	fMain(0);
#endif

	return 0;
}