#include <bits/stdc++.h>
using namespace std;
int n;
unsigned long long h[1100000];
unsigned long long power[1100000];
char s[1100000];
int r[1100000];
unsigned long long query(int L, int R) {
  return (h[R] - h[L - 1] * power[R - L + 1] % 1000000007 + 1000000007) %
         1000000007;
}
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  power[0] = 1;
  for (int i = 1; i <= n; i++) {
    h[i] = (h[i - 1] * 107 + s[i]) % 1000000007;
    power[i] = (power[i - 1] * 107) % 1000000007;
  }
  for (int i = 1; i <= n / 2; i++) {
    int L = 1, R = i;
    if (s[i] != s[n + 1 - i]) continue;
    while (L != R) {
      int mid = (L + R + 1) / 2;
      if (query(i - mid + 1, i + mid - 1) ==
          query(n + 1 - i - mid + 1, n + 1 - i + mid - 1))
        L = mid;
      else
        R = mid - 1;
    }
    r[i - L + 1] = max(r[i - L + 1], i);
  }
  for (int i = 1, j = 0; i <= n / 2; i++) {
    j = max(j, r[i]);
    if (j < i)
      printf("%d ", -1);
    else
      printf("%d ", (j - i) * 2 + 1);
  }
  if (n & 1) printf("-1");
  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;
int n;
unsigned long long h[1100000];
unsigned long long power[1100000];
char s[1100000];
int r[1100000];
unsigned long long query(int L, int R) {
  return (h[R] - h[L - 1] * power[R - L + 1] % 1000000007 + 1000000007) %
         1000000007;
}
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  power[0] = 1;
  for (int i = 1; i <= n; i++) {
    h[i] = (h[i - 1] * 107 + s[i]) % 1000000007;
    power[i] = (power[i - 1] * 107) % 1000000007;
  }
  for (int i = 1; i <= n / 2; i++) {
    int L = 1, R = i;
    if (s[i] != s[n + 1 - i]) continue;
    while (L != R) {
      int mid = (L + R + 1) / 2;
      if (query(i - mid + 1, i + mid - 1) ==
          query(n + 1 - i - mid + 1, n + 1 - i + mid - 1))
        L = mid;
      else
        R = mid - 1;
    }
    r[i - L + 1] = max(r[i - L + 1], i);
  }
  for (int i = 1, j = 0; i <= n / 2; i++) {
    j = max(j, r[i]);
    if (j < i)
      printf("%d ", -1);
    else
      printf("%d ", (j - i) * 2 + 1);
  }
  if (n & 1) printf("-1");
  return 0;
}

```