#include <bits/stdc++.h>
using namespace std;
const int N = 4e6 + 10;
char str[N], s[N];
int f[N], ans[N];
int main() {
  int n;
  scanf("%d%s", &n, str);
  for (int i = 1; i <= n * 4 + 1; i++) {
    if (i & 1)
      s[i] = '#';
    else if (i >> 1 & 1)
      s[i] = str[i >> 2];
    else
      s[i] = str[n - (i >> 2)];
  }
  int mid = 0, r = 0;
  for (int i = 1; i <= n * 4 + 1; i++) {
    f[i] = i < r ? min(r - i, f[mid * 2 - i]) : 1;
    while (f[i] < i && i + f[i] <= n * 4 + 1 && s[i - f[i]] == s[i + f[i]])
      f[i]++;
    if (f[i] + i - 1 > r) r = f[i] + i - 1, mid = i;
  }
  for (int i = 1; i <= n * 2; i++) ans[i] = -1;
  for (int i = 1; i <= n * 2; i++)
    if (i & 1) {
      if (f[i] == 1) continue;
      if (f[i] >> 1 & 1 ^ 1) f[i] -= 2;
      if (f[i] >> 1 & 1)
        ans[(i - f[i] + 2) >> 1] = max(ans[(i - f[i] + 2) >> 1], f[i] >> 1);
    }
  for (int i = 2; i <= n; i++) ans[i] = max(ans[i], ans[i - 2] - 2);
  for (int i = 1; i <= n; i += 2) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 = 4e6 + 10;
char str[N], s[N];
int f[N], ans[N];
int main() {
  int n;
  scanf("%d%s", &n, str);
  for (int i = 1; i <= n * 4 + 1; i++) {
    if (i & 1)
      s[i] = '#';
    else if (i >> 1 & 1)
      s[i] = str[i >> 2];
    else
      s[i] = str[n - (i >> 2)];
  }
  int mid = 0, r = 0;
  for (int i = 1; i <= n * 4 + 1; i++) {
    f[i] = i < r ? min(r - i, f[mid * 2 - i]) : 1;
    while (f[i] < i && i + f[i] <= n * 4 + 1 && s[i - f[i]] == s[i + f[i]])
      f[i]++;
    if (f[i] + i - 1 > r) r = f[i] + i - 1, mid = i;
  }
  for (int i = 1; i <= n * 2; i++) ans[i] = -1;
  for (int i = 1; i <= n * 2; i++)
    if (i & 1) {
      if (f[i] == 1) continue;
      if (f[i] >> 1 & 1 ^ 1) f[i] -= 2;
      if (f[i] >> 1 & 1)
        ans[(i - f[i] + 2) >> 1] = max(ans[(i - f[i] + 2) >> 1], f[i] >> 1);
    }
  for (int i = 2; i <= n; i++) ans[i] = max(ans[i], ans[i - 2] - 2);
  for (int i = 1; i <= n; i += 2) printf("%d ", ans[i]);
  return 0;
}

```