Bill is a famous mathematician in BubbleLand. Thanks to his revolutionary math discoveries he was able to make enough money to build a beautiful house. Unfortunately, for not paying property tax on time, court decided to punish Bill by making him lose a part of his property.

Bill’s property can be observed as a convex regular 2n-sided polygon A0 A1... A2n - 1 A2n, A2n = A0, with sides of the exactly 1 meter in length. 

Court rules for removing part of his property are as follows:

  * Split every edge Ak Ak + 1, k = 0... 2n - 1 in n equal parts of size 1 / n with points P0, P1, ..., Pn - 1
  * On every edge A2k A2k + 1, k = 0... n - 1 court will choose one point B2k = Pi for some i = 0, ..., n - 1 such that <image>
  * On every edge A2k + 1A2k + 2, k = 0...n - 1 Bill will choose one point B2k + 1 = Pi for some i = 0, ..., n - 1 such that <image>
  * Bill gets to keep property inside of 2n-sided polygon B0 B1... B2n - 1



Luckily, Bill found out which B2k points the court chose. Even though he is a great mathematician, his house is very big and he has a hard time calculating. Therefore, he is asking you to help him choose points so he maximizes area of property he can keep.

Input

The first line contains one integer number n (2 ≤ n ≤ 50000), representing number of edges of 2n-sided polygon.

The second line contains n distinct integer numbers B2k (0 ≤ B2k ≤ n - 1, k = 0... n - 1) separated by a single space, representing points the court chose. If B2k = i, the court chose point Pi on side A2k A2k + 1.

Output

Output contains n distinct integers separated by a single space representing points B1, B3, ..., B2n - 1 Bill should choose in order to maximize the property area. If there are multiple solutions that maximize the area, return any of them.

Example

Input

3
0 1 2


Output

0 2 1

Note

To maximize area Bill should choose points: B1 = P0, B3 = P2, B5 = P1

<image>
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <typename T>
T lcm(T a, T b) {
  return (a / gcd(a, b)) * b;
}
string int_to_str_easy(long long n) {
  return n ? "" : int_to_str_easy(n / 10) + (char)('0' + n % 10);
}
template <typename T>
bool is_prime_sqrt(T n) {
  if (n == 1) return false;
  T k = trunc(sqrt(n));
  for (T i = 2; i <= k; i++)
    if (n % i == 0) return false;
  return true;
}
bool is_down(char x) { return ('a' <= x && x <= 'z'); }
bool is_upper(char x) { return ('A' <= x && x <= 'Z'); }
bool is_digit(char x) { return ('0' <= x && x <= '9'); }
string follow(string s) {
  for (int i = 0; i < s.length(); i++)
    if (is_upper(s[i])) s[i] += 'a' - 'A';
  return s;
}
const long double pi = 3.141592653589793238462643383279;
const long double log23 = 1.58496250072115618145373894394781;
const long double eps = 1e-8;
const long double zero = 0;
const long long INF = 1e18;
const int COUT = 30;
const int prost = 239;
const long long prost64 = 239;
const int two = 2;
const long long sr = 31;
const long long MOD = 1000000007;
const int BIG = 1e9;
const int alf = 26;
const int MAX_N = 1e5 + 1;
const int MAX_T = (1 << 20);
const int BLOCK = trunc(sqrt(MAX_N)) + 1;
const int MAX_LOG = 21;
const int km = (1 << 18);
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const int bit_max = 32;
const int dig = 10;
const string str_alf = "abcdefghijklmnopqrstuvwxyz";
const string str_alf_big = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n, a[MAX_N];
int main() {
  cin.sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<pair<int, int> > v;
  for (int i = 0; i < n; i++) {
    int p1 = n - a[i];
    int p2 = a[(i + 1) % n];
    v.push_back(make_pair(p2 - p1, i));
  }
  sort(v.begin(), v.end());
  vector<int> ans(n);
  for (int i = 0; i < n; i++) ans[v[i].second] = i;
  for (int i : ans) cout << i << " ";
  return 0;
}
