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;
const int maxn = 5e4 + 5;
const double pi = acosl(-1.0);
struct Vec {
  double x, y;
  Vec() {}
  Vec(double a, double b) : x(a), y(b) {}
  Vec operator+(const Vec &t) const { return Vec(x + t.x, y + t.y); }
  Vec operator-(const Vec &t) const { return Vec(x - t.x, y - t.y); }
} point[4][maxn];
double cro(Vec a, Vec b) { return a.x * b.y - a.y * b.x; }
pair<int, int> query[maxn];
int arr[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  double alpth = pi / n;
  Vec now = Vec(0, 0), dir = Vec(10, 0);
  for (int o = 1; o <= 3; o++) {
    for (int i = 1; i <= n; i++) {
      point[o][i] = now;
      now = now + dir;
    }
    dir = Vec(10 * cos(alpth), 10 * sin(alpth));
    alpth += pi / n;
  }
  arr[n] = arr[0];
  for (int i = 0; i < n; i++) {
    Vec base = point[1][arr[i] + 1];
    Vec to = point[3][arr[i + 1] + 1];
    double area1 = cro(point[2][1] - base, to - base);
    double area2 = cro(point[2][2] - base, to - base);
    query[i] = {arr[i] + arr[i + 1], i};
  }
  sort(query, query + n);
  vector<int> ans(n);
  for (int i = 0; i < n; i++) ans[query[i].second] = i;
  for (int i = 0; i < n; i++) {
    printf("%d", ans[i]);
    if (i != n) printf(" ");
  }
  puts("");
  return 0;
}
