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)

#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<cstdio>
#include<bitset>
#include<unordered_map>
using namespace std;
//---------------------------------------------------
//ライブラリゾーン！！！！
typedef long long ll;
typedef long double ld;
#define str string
#define rep(i,j) for(ll i=0;i<(long long)(j);i++)
const ll Mod = 1000000007;
const ll gosenchou = 5000000000000000;
short gh[2][4] = { { 0,0,-1,1 },{ -1,1,0,0 } };
struct P {
	ll pos, cost;
};
bool operator<(P a, P b) { return a.cost < b.cost; }
bool operator>(P a, P b) { return a.cost > b.cost; }
struct B {//隣接リスト表現
	ll to;
	ld cost;
};
struct E {//辺の情報を入れる変数
	ll from, to, cost;
};
bool operator<(E a, E b) {
	return a.cost < b.cost;
}
struct H {
	ll x, y;
};
bool operator<(H a, H b) {
	/*if (a.x != b.x) return a.x < b.x;
	return a.y < b.y;*/
	return ((a.x + 1)*(a.y + 1)) < ((b.x + 1)*(b.y + 1));
}
bool operator>(H a, H b) {
	if (a.x != b.x) return a.x > b.x;
	return a.y > b.y;
}
bool operator==(H a, H b) {
	return a.x == b.x&&a.y == b.y;
}
bool operator!=(H a, H b) {
	return a.x != b.x || a.y != b.y;
}
ll gcm(ll i, ll j) {//最大公約数
	if (i > j) swap(i, j);
	if (i == 0) return j;
	return gcm(j%i, i);
}
ld rad(H a, H b) {
	return sqrt(pow(a.x - b.x, 2.0) + pow(a.y - b.y, 2.0));
}//rad＝座標上の2点間の距離
ll ari(ll a, ll b, ll c) {
	return (a + b)*c / 2;
}//等差数列の和
ll fact(ll x, ll k, ll p) {//最大値、個数
	ll sum = 1;
	for (int i = 0; i < k; i++) {
		sum *= (x--);
		sum %= p;
	}
	return sum;
}//階乗(正）
ll mod_pow(ll x, ll n, ll p) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res*x%p;
		x = x*x%p;
		n >>= 1;
	}
	return res;
}//x^n%p
int ctoi(char a) {
	return (int)a - '0';
}
 //#define int long long
const long long Inf = 4523372036854775807;
const int inf = 1500000000;
//----------------------------------------------------
//++++++++++++++++++++++++++++++++++++++++++++++++++++
int n;
int a[100002];
signed main() {
	cin >> n;
	n++;
	for (int i = 1; i < n; i++) {
		cin >> a[i];
	}
	int ans = 0;
	for (int i = 1; i <= n; i++) {
		ans += abs(a[i] - a[i - 1]);
	}
	for (int i = 1; i < n; i++) {
		if (a[i - 1]<a[i]&&a[i]<a[i+1]) {
			cout << ans << endl;
		}
		else if (a[i + 1] < a[i] && a[i] < a[i - 1]) {
			cout << ans << endl;
		}
		else{
			cout << ans - abs(a[i] - a[i - 1]) - abs(a[i] - a[i + 1]) + abs(a[i - 1] - a[i + 1]) << endl;
		}
	}
	getchar(); getchar();
}