#include <bits/stdc++.h>
using namespace std;
int n, A[1000010];
char s[1000010];
long long h[1000010], bas[1000010];
inline long long gH(int l, int r) {
  return (h[r] - h[l - 1] * bas[r - l + 1] % 1000000007 + 1000000007) %
         1000000007;
}
int main() {
  scanf("%d%s", &n, s + 1);
  for (int i = *bas = 1; i <= n; ++i) {
    bas[i] = bas[i - 1] * 131 % 1000000007;
    h[i] = (h[i - 1] * 131 + s[i] - 'a') % 1000000007;
  }
  memset(A, -1, sizeof A);
  for (int i = n + 1 >> 1, j; i; --i) {
    j = n - i + 1;
    if (i == j)
      A[i] = -1;
    else
      for (A[i] = A[i + 1] + 2; ~A[i]; A[i] -= 2)
        if (gH(i, i + A[i] - 1) == gH(j - A[i] + 1, j)) break;
  }
  for (int i = 1; i <= n + 1 >> 1; ++i) printf("%d ", A[i]);
}

### Prompt

Generate 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;
int n, A[1000010];
char s[1000010];
long long h[1000010], bas[1000010];
inline long long gH(int l, int r) {
  return (h[r] - h[l - 1] * bas[r - l + 1] % 1000000007 + 1000000007) %
         1000000007;
}
int main() {
  scanf("%d%s", &n, s + 1);
  for (int i = *bas = 1; i <= n; ++i) {
    bas[i] = bas[i - 1] * 131 % 1000000007;
    h[i] = (h[i - 1] * 131 + s[i] - 'a') % 1000000007;
  }
  memset(A, -1, sizeof A);
  for (int i = n + 1 >> 1, j; i; --i) {
    j = n - i + 1;
    if (i == j)
      A[i] = -1;
    else
      for (A[i] = A[i + 1] + 2; ~A[i]; A[i] -= 2)
        if (gH(i, i + A[i] - 1) == gH(j - A[i] + 1, j)) break;
  }
  for (int i = 1; i <= n + 1 >> 1; ++i) printf("%d ", A[i]);
}

```