cf #314 A. Lineland Mail

All cities of Lineland are located on the Ox coordinate axis. Thus, each city is associated with its position xi — a coordinate on the Ox axis. No two cities are located at a single point.

Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).

Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.

For each city calculate two values mini and maxi, where mini is the minimum cost of sending a letter from the i-th city to some other city, and maxi is the the maximum cost of sending a letter from the i-th city to some other city

Input

The first line of the input contains integer n (2 ≤ n ≤ 105) — the number of cities in Lineland. The second line contains the sequence of ndistinct integers x1, x2, ..., xn ( - 109 ≤ xi ≤ 109), where xi is the x-coordinate of the i-th city. All the xi's are distinct and follow inascending order.

Output

Print n lines, the i-th line must contain two integers mini, maxi, separated by a space, where mini is the minimum cost of sending a letter from the i-th city, and maxi is the maximum cost of sending a letter from the i-th city.

Sample test(s)

input

4
-5 -2 2 7

output

input

2
-1 1

output

2 2
2 2

给一个有序序列，问对于没一个数，和它相差最少和最多的数的位置。

/*************************************************************************
    > File Name: code/cf/#314/A.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月06日 星期四 00时01分51秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=2E5+7;
LL a[N];
int main()
{
    int n;
    cin>>n;
    a[0]=-inf;
    a[n+1]=inf;

    for ( int i  = 1 ; i <= n ; i ++ )
    {
    cin>>a[i];
    }
    int mx = -1;
    int mi = inf;
    cout<<a[2]-a[1]<<" "<<a[n]-a[1]<<endl;
    for ( int i = 2 ; i <= n-1 ; i++ )
    {
    cout<<min(abs(a[i]-a[i-1]),abs(a[i+1]-a[i]))<<" "<<max(abs(a[i]-a[1]),abs(a[n]-a[i]))<<endl;
    }
    cout<<a[n]-a[n-1]<<" "<<a[n]-a[1]<<endl;
    return 0;
}