#include <bits/stdc++.h>
char s[1000100];
int hs[1000100], pw[1000100], n;
int ans[1000100];
int hash(int l, int r) {
  return (hs[r] - ((long long)pw[r - l + 1] * hs[l - 1]) % 1000000007 +
          1000000007) %
         1000000007;
}
void init() {
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = ((long long)pw[i - 1] * 26) % 1000000007;
  for (int i = 1; i <= n; i++)
    hs[i] = ((long long)hs[i - 1] * 26 + s[i] - 'a') % 1000000007;
  for (int i = 1; i <= (n + 1) / 2; i++) ans[i] = -1;
}
int main() {
  scanf("%d%s", &n, s + 1);
  init();
  if (n % 2)
    ans[(n + 1) / 2] = -1;
  else
    ans[n / 2] = (s[n / 2] == s[n / 2 + 1] ? 1 : -1);
  for (int i = (n - 1) / 2; i >= 1; i--) {
    for (int j = ans[i + 1] + 2; j >= 1; j -= 2) {
      if (hash(i, i + j - 1) == hash(n + 2 - i - j, n + 1 - i)) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= (n + 1) / 2; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

### Prompt

Create a solution in CPP for 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>
char s[1000100];
int hs[1000100], pw[1000100], n;
int ans[1000100];
int hash(int l, int r) {
  return (hs[r] - ((long long)pw[r - l + 1] * hs[l - 1]) % 1000000007 +
          1000000007) %
         1000000007;
}
void init() {
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = ((long long)pw[i - 1] * 26) % 1000000007;
  for (int i = 1; i <= n; i++)
    hs[i] = ((long long)hs[i - 1] * 26 + s[i] - 'a') % 1000000007;
  for (int i = 1; i <= (n + 1) / 2; i++) ans[i] = -1;
}
int main() {
  scanf("%d%s", &n, s + 1);
  init();
  if (n % 2)
    ans[(n + 1) / 2] = -1;
  else
    ans[n / 2] = (s[n / 2] == s[n / 2 + 1] ? 1 : -1);
  for (int i = (n - 1) / 2; i >= 1; i--) {
    for (int j = ans[i + 1] + 2; j >= 1; j -= 2) {
      if (hash(i, i + j - 1) == hash(n + 2 - i - j, n + 1 - i)) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= (n + 1) / 2; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

```