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>
#define sz(x) (int)(x).size()
#define For(i, a, b) for (int i = int(a); i < int(b); i++)
#define Dor(i, a, b) for (int i = int(b) - 1; i >= int(a); i--)
#define Foz(i, a) For(i, 0, a)
#define Doz(i, a) Dor(i, 0, a)
#define Trav(e, v) for (auto& e : v)
#define pb push_back
#define mp make_pair
#define all(x) begin(x), end(x)
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vii = vector<ii>;
template <typename T> bool ckmin (T& a, T b) { return (b < a) ? a = b, 1 : 0; }
template <typename T> bool ckmax (T& a, T b) { return (b > a) ? a = b, 1 : 0; }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#ifdef XVENOM
	#define errp(...) fprintf(stderr, __VA_ARGS__)
	#define dbg(...) errp("[%s]: ", #__VA_ARGS__), dbg_str(__VA_ARGS__)
#else
	#define errp(...)
	#define dbg(...)
#endif
template <typename U, typename V> string to_string (pair<U, V>);
string to_string (string& e) { return "\"" + e + "\""; }
string to_string (char e) { return "\'" + string(1, e) + "\'"; }
string to_string (bool e) { return e ? "true" : "false"; }
template <typename T> string to_string (T e) {
	string s = "[ ";
	for (auto& x : e) s += to_string(x) + " ";
	return s + "]";
}
template <typename U, typename V> string to_string (pair<U, V> e) {
	return "(" + to_string(e.first) + ", " + to_string(e.second) + ")";
}
void dbg_str () { errp(" \n"); }
template <typename U, typename... V> void dbg_str (U u, V... v) {
 	errp(" %s", to_string(u).c_str()); dbg_str(v...);
}
/* --- */
const int N = 100010;
int a[N], p[N], s[N], n;

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);
	/* --- */
	cin >> n;
	For(i, 1, n + 1) cin >> a[i];
	a[0] = a[n + 1] = p[0] = s[n + 1] = 0;
	For(i, 1, n + 1) p[i] = p[i - 1] + abs(a[i] - a[i - 1]);
	Dor(i, 1, n + 1) s[i] = s[i + 1] + abs(a[i] - a[i + 1]);
	For(i, 1, n + 1) cout << p[i - 1] + s[i + 1] + abs(a[i - 1] - a[i + 1]) << '\n';
}