#include <bits/stdc++.h>
using namespace std;
static const int b = 43;
static const int mod = 1000000007;
int main() {
  int n;
  scanf("%d", &n);
  char *s = (char *)malloc((n + 1) * sizeof(char));
  scanf("%s", s);
  vector<long long> h(n), pw(n);
  for (int i = 0; i < n; ++i) {
    h[i] = ((i ? 1LL * h[i - 1] * b : 0) + s[i] - 'a') % mod;
    pw[i] = i ? (1LL * pw[i - 1] * b % mod) : 1;
  }
  vector<int> ans((n + 1) / 2, -1);
  for (int i = 0; i < ans.size(); ++i) {
    int j = n - i - 1;
    if (i >= j || s[i] != s[j]) continue;
    int lower = 0, upper = i;
    while (lower < upper) {
      int mid = (lower + upper + 1) / 2;
      long long h1 = h[i + mid];
      if (i > mid)
        h1 = (h1 - 1LL * h[i - mid - 1] * pw[mid * 2 + 1] % mod + mod) % mod;
      long long h2 = h[j + mid];
      if (j > mid)
        h2 = (h2 - 1LL * h[j - mid - 1] * pw[mid * 2 + 1] % mod + mod) % mod;
      if (h1 == h2)
        lower = mid;
      else
        upper = mid - 1;
    }
    ans[i - lower] = max(ans[i - lower], 2 * lower + 1);
  }
  for (int i = 0; i < ans.size(); ++i) {
    if (i) ans[i] = max(ans[i], ans[i - 1] - 2);
    printf("%d%c", ans[i], " \n"[i + 1 == ans.size()]);
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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;
static const int b = 43;
static const int mod = 1000000007;
int main() {
  int n;
  scanf("%d", &n);
  char *s = (char *)malloc((n + 1) * sizeof(char));
  scanf("%s", s);
  vector<long long> h(n), pw(n);
  for (int i = 0; i < n; ++i) {
    h[i] = ((i ? 1LL * h[i - 1] * b : 0) + s[i] - 'a') % mod;
    pw[i] = i ? (1LL * pw[i - 1] * b % mod) : 1;
  }
  vector<int> ans((n + 1) / 2, -1);
  for (int i = 0; i < ans.size(); ++i) {
    int j = n - i - 1;
    if (i >= j || s[i] != s[j]) continue;
    int lower = 0, upper = i;
    while (lower < upper) {
      int mid = (lower + upper + 1) / 2;
      long long h1 = h[i + mid];
      if (i > mid)
        h1 = (h1 - 1LL * h[i - mid - 1] * pw[mid * 2 + 1] % mod + mod) % mod;
      long long h2 = h[j + mid];
      if (j > mid)
        h2 = (h2 - 1LL * h[j - mid - 1] * pw[mid * 2 + 1] % mod + mod) % mod;
      if (h1 == h2)
        lower = mid;
      else
        upper = mid - 1;
    }
    ans[i - lower] = max(ans[i - lower], 2 * lower + 1);
  }
  for (int i = 0; i < ans.size(); ++i) {
    if (i) ans[i] = max(ans[i], ans[i - 1] - 2);
    printf("%d%c", ans[i], " \n"[i + 1 == ans.size()]);
  }
  return 0;
}

```