#include <bits/stdc++.h>
using namespace std;
class RabinKarp {
  long long m;
  vector<int> pw, hsh;

 public:
  RabinKarp() {}
  RabinKarp(char str[], long long p, long long _m) : m(_m) {
    int n = strlen(str);
    pw.resize(n);
    hsh.resize(n);
    hsh[0] = str[0];
    pw[0] = 1;
    for (int i = 1; i < n; i++) {
      hsh[i] = (hsh[i - 1] * p + str[i]) % m;
      pw[i] = (pw[i - 1] * p) % m;
    }
  }
  long long hash(int i, int j) {
    long long ans = hsh[j];
    if (i > 0) ans = (ans - ((hsh[i - 1] * 1ll * pw[j - i + 1]) % m) + m) % m;
    return ans;
  }
};
char str[1000009];
int N, f[1000009];
RabinKarp rk;
bool check(int l, int r) {
  if (f[l] <= 0) return true;
  if (l + f[l] - 1 >= r) return false;
  return rk.hash(l, l + f[l] - 1) == rk.hash(r - f[l] + 1, r);
}
int main() {
  scanf("%d %s", &N, str);
  rk = RabinKarp(str, 256, 1000000007);
  int r = N / 2;
  int l = (N % 2 == 0 ? r - 1 : r);
  int k = l;
  f[l + 1] = -1;
  for (; l >= 0; l--, r++) {
    f[l] = f[l + 1] + 2;
    while (!check(l, r)) f[l] -= 2;
  }
  for (int i = 0; i <= k; i++) printf("%d ", f[i]);
  printf("\n");
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

You are given a string s consisting of n lowercase Latin letters.

Let's denote k-substring of s as a string subsk = sksk + 1..sn + 1 - k. Obviously, subs1 = s, and there are exactly <image> such substrings.

Let's call some string t an odd proper suprefix of a string T iff the following conditions are met:

  * |T| > |t|; 
  * |t| is an odd number; 
  * t is simultaneously a prefix and a suffix of T.



For evey k-substring (<image>) of s you have to calculate the maximum length of its odd proper suprefix.

Input

The first line contains one integer n (2 ≤ n ≤ 106) — the length s.

The second line contains the string s consisting of n lowercase Latin letters.

Output

Print <image> integers. i-th of them should be equal to maximum length of an odd proper suprefix of i-substring of s (or  - 1, if there is no such string that is an odd proper suprefix of i-substring).

Examples

Input

15
bcabcabcabcabca


Output

9 7 5 3 1 -1 -1 -1


Input

24
abaaabaaaabaaabaaaabaaab


Output

15 13 11 9 7 5 3 1 1 -1 -1 1


Input

19
cabcabbcabcabbcabca


Output

5 3 1 -1 -1 1 1 -1 -1 -1

Note

The answer for first sample test is folowing: 

  * 1-substring: bcabcabcabcabca
  * 2-substring: cabcabcabcabc
  * 3-substring: abcabcabcab
  * 4-substring: bcabcabca
  * 5-substring: cabcabc
  * 6-substring: abcab 
  * 7-substring: bca 
  * 8-substring: c 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
class RabinKarp {
  long long m;
  vector<int> pw, hsh;

 public:
  RabinKarp() {}
  RabinKarp(char str[], long long p, long long _m) : m(_m) {
    int n = strlen(str);
    pw.resize(n);
    hsh.resize(n);
    hsh[0] = str[0];
    pw[0] = 1;
    for (int i = 1; i < n; i++) {
      hsh[i] = (hsh[i - 1] * p + str[i]) % m;
      pw[i] = (pw[i - 1] * p) % m;
    }
  }
  long long hash(int i, int j) {
    long long ans = hsh[j];
    if (i > 0) ans = (ans - ((hsh[i - 1] * 1ll * pw[j - i + 1]) % m) + m) % m;
    return ans;
  }
};
char str[1000009];
int N, f[1000009];
RabinKarp rk;
bool check(int l, int r) {
  if (f[l] <= 0) return true;
  if (l + f[l] - 1 >= r) return false;
  return rk.hash(l, l + f[l] - 1) == rk.hash(r - f[l] + 1, r);
}
int main() {
  scanf("%d %s", &N, str);
  rk = RabinKarp(str, 256, 1000000007);
  int r = N / 2;
  int l = (N % 2 == 0 ? r - 1 : r);
  int k = l;
  f[l + 1] = -1;
  for (; l >= 0; l--, r++) {
    f[l] = f[l + 1] + 2;
    while (!check(l, r)) f[l] -= 2;
  }
  for (int i = 0; i <= k; i++) printf("%d ", f[i]);
  printf("\n");
  return 0;
}

```