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 maxx = 600010;
const double pi = acos(-1.0);
const double eps = 1e-15;
struct node {
  int num, val;
  bool operator<(const node &X) const { return val < X.val; }
};
multiset<node> s;
multiset<node>::iterator it;
int t, n, m;
int ans = 0, total = 0, sum = 0, cnt = 0;
int a[maxx], b[maxx];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[n + 1] = a[1];
  for (int i = 1; i <= n; i++) s.insert(node{i, a[i] + a[i + 1]});
  total = 0;
  for (it = s.begin(); it != s.end(); it++) {
    b[it->num] = total;
    total++;
  }
  for (int i = 1; i <= n; i++) printf("%d ", b[i]);
  return 0;
}
