#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
const int base = 131;
const int mod = 1e9 + 7;
char s[N];
int n, ans[N / 2];
long long h[N], p[N];
long long get(int l, int r) {
  return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
}
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  p[0] = 1;
  for (int i = 1; i <= n; i++) p[i] = p[i - 1] * base % mod;
  for (int i = 1; i <= n; i++)
    h[i] = (h[i - 1] * base % mod + s[i] - 'a') % mod;
  int l, r;
  int mid = n + 1 >> 1;
  if (n % 2 == 0) {
    l = n >> 1, r = l + 1;
    if (s[l] == s[r])
      ans[mid] = 1;
    else
      ans[mid] = -1;
  } else
    ans[mid] = -1, l = r = mid;
  for (int i = mid - 1; i >= 1; i--) {
    l--, r++;
    for (int j = ans[i + 1] + 2; j >= -1; j -= 2) {
      if (get(l, l + j - 1) == get(r - j + 1, r) || j == -1) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= mid; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int N = 2e6 + 10;
const int base = 131;
const int mod = 1e9 + 7;
char s[N];
int n, ans[N / 2];
long long h[N], p[N];
long long get(int l, int r) {
  return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
}
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  p[0] = 1;
  for (int i = 1; i <= n; i++) p[i] = p[i - 1] * base % mod;
  for (int i = 1; i <= n; i++)
    h[i] = (h[i - 1] * base % mod + s[i] - 'a') % mod;
  int l, r;
  int mid = n + 1 >> 1;
  if (n % 2 == 0) {
    l = n >> 1, r = l + 1;
    if (s[l] == s[r])
      ans[mid] = 1;
    else
      ans[mid] = -1;
  } else
    ans[mid] = -1, l = r = mid;
  for (int i = mid - 1; i >= 1; i--) {
    l--, r++;
    for (int j = ans[i + 1] + 2; j >= -1; j -= 2) {
      if (get(l, l + j - 1) == get(r - j + 1, r) || j == -1) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= mid; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}

```