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 = 100010;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const double PI = acos(-1);
template <typename T>
inline void read(T &x) {
  x = 0;
  T f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do x = x * 10 + ch - '0', ch = getchar();
  while (ch <= '9' && ch >= '0');
  x *= f;
}
template <typename A, typename B>
inline void read(A &x, B &y) {
  read(x);
  read(y);
}
template <typename A, typename B, typename C>
inline void read(A &x, B &y, C &z) {
  read(x);
  read(y);
  read(z);
}
template <typename A, typename B, typename C, typename D>
inline void read(A &x, B &y, C &z, D &w) {
  read(x);
  read(y);
  read(z);
  read(w);
}
template <typename A, typename B>
inline A fexp(A x, B p) {
  A ans = 1;
  for (; p; p >>= 1, x = 1LL * x * x % mod)
    if (p & 1) ans = 1LL * ans * x % mod;
  return ans;
}
template <typename A, typename B>
inline A fexp(A x, B p, A mo) {
  A ans = 1;
  for (; p; p >>= 1, x = 1LL * x * x % mo)
    if (p & 1) ans = 1LL * ans * x % mo;
  return ans;
}
int n;
int A[maxn], B[maxn];
pair<int, int> Ans[maxn];
void Work() {
  for (int i = (0); i <= (n - 1); i++) printf("%d ", B[i]);
}
void Init() {
  read(n);
  for (int i = (0); i <= (n - 1); i++) read(A[i]);
  for (int i = (0); i <= (n - 1); i++)
    Ans[i] = make_pair(A[(i + 1) % n] - (n - A[i]), i);
  sort(Ans, Ans + n);
  for (int i = (0); i <= (n - 1); i++) B[Ans[i].second] = i;
}
int main() {
  Init();
  Work();
  return 0;
}
