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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <cmath>
#include <numeric>
using namespace std;
#define rep(i,a,b) for(int64_t i=(a); i<(b); ++i) // a ≦ i < b 
#define Rrep(i,a,b) for(int64_t i=(a);i>=(b);--i) // reverse repeat. a から b まで減少．
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend() //逆イテレータ
#define RANGE(a,b,c) (a).begin()+b,(a).begin()+c // [b,c) イテレータ
#define MOD 1000000007
#define INF 1000000000
typedef pair<int64_t, int64_t> PII;
typedef vector<int64_t> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef vector<PII> VP;

int main() {
cin.tie(0);
ios::sync_with_stdio(false);

int64_t N; cin >> N;
VI A(N); rep(i, 0, N) cin >> A[i];
int64_t all = abs(A[0]) + abs(A[N-1]); 
rep(i, 0, N-1) all += abs(A[i] - A[i+1]);

cout << all - abs(A[0]) - abs(A[0] - A[1]) + abs(A[1]) << "\n";
rep(i, 1, N-1)
    cout << all - abs(A[i-1] - A[i]) - abs(A[i] - A[i+1]) + abs(A[i-1] - A[i+1]) << "\n";
cout << all - abs(A[N-2] - A[N-1]) - abs(A[N-1]) + abs(A[N-2]) << "\n";


}
// 書いて考える
// 境界，出力文字列 チェック
// 可読性優先．高速化次点．
// まずは全探索，分割統治，次にDP
// 制限を見る．境界に注意．求めたい量の変域．
// 偶奇，逆から，ソート，出現回数，出現位置，DP， 余事象，包除
// データ構造． 問題の特徴量．単調性，二分探索
// 存在判定：構成方法，入力の特徴
// gcd, lcm ,素因数分解．
// 例外を十分に含む一般化．想像力の限界
// 小さい系から例示
// 代数的処理
// 始めは過剰に例示・場合分けしてもいい．各場合を確実に対処．
// 自明な例から処理，除外．
// 小数のときは，精度の設定する．doubel 変数に数値を入力するときは 123. とする．
// テストケース作成は数表あり