#include <bits/stdc++.h>
using std::set;
int n, po[1000010], A[1000010], ans[1000010];
inline int a(int x, int y) {
  return ((A[y] - 1LL * A[x - 1] * po[y - x + 1] % 1061109559) + 1061109559) %
         1061109559;
}
char buf[1000010];
int main() {
  scanf("%d%s", &n, buf + 1);
  for (int i = po[0] = 1; i <= n; i++)
    po[i] = 1LL * po[i - 1] * 37 % 1061109559;
  for (int i = 1; i <= n; i++)
    A[i] = (1LL * A[i - 1] * 37 + buf[i] - 'a') % 1061109559;
  for (int i = (n + 1) >> 1, now = -1; i; i--) {
    now += 2;
    if (i * 2 > n) now = -1;
    while (~now && a(i, i + now - 1) != a(n - i - now + 2, n - i + 1)) now -= 2;
    ans[i] = now;
  }
  for (int i = 1; i <= (n + 1) >> 1; i++)
    printf("%d%c", ans[i], " \n"[i == (n + 1) >> 1]);
  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>
using std::set;
int n, po[1000010], A[1000010], ans[1000010];
inline int a(int x, int y) {
  return ((A[y] - 1LL * A[x - 1] * po[y - x + 1] % 1061109559) + 1061109559) %
         1061109559;
}
char buf[1000010];
int main() {
  scanf("%d%s", &n, buf + 1);
  for (int i = po[0] = 1; i <= n; i++)
    po[i] = 1LL * po[i - 1] * 37 % 1061109559;
  for (int i = 1; i <= n; i++)
    A[i] = (1LL * A[i - 1] * 37 + buf[i] - 'a') % 1061109559;
  for (int i = (n + 1) >> 1, now = -1; i; i--) {
    now += 2;
    if (i * 2 > n) now = -1;
    while (~now && a(i, i + now - 1) != a(n - i - now + 2, n - i + 1)) now -= 2;
    ans[i] = now;
  }
  for (int i = 1; i <= (n + 1) >> 1; i++)
    printf("%d%c", ans[i], " \n"[i == (n + 1) >> 1]);
  return 0;
}

```