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 <int n>
using AI = array<int, n>;
const double eps = 1e-9;
const int oo = 0x3F3F3F3F;
const long long ooLL = 0x3F3F3F3F3F3F3F3FLL;
const int MOD = 1000000007;
template <typename T>
T sqr(T x) {
  return x * x;
}
const double PI = acos(-1.);
long long Round(double x) { return x < 0 ? x - .5 : x + .5; }
int main() {
  void run();
  run();
  return 0;
}
int n, b[1 << 17];
pair<int, int> e[1 << 17];
void run() {
  scanf("%d", &n);
  for (int i = (0), _b(n); i < _b; ++i) scanf("%d", b + i);
  for (int i = (0), _b(n); i < _b; ++i) {
    int p = n - b[i], q = b[(i + 1) % n];
    e[i] = make_pair(q - p, i);
  }
  sort(e, e + n);
  for (int i = (0), _b(n); i < _b; ++i) b[e[i].second] = i;
  for (int i = (0), _b(n); i < _b; ++i) printf("%d%c", b[i], "\n "[i + 1 < n]);
}
