#include <bits/stdc++.h>
using namespace std;
char s[1000005];
int n, h[1000005], pw[1000005], ans[1000005], l, r, len;
int gethash(int s, int t) {
  return (h[t] + 20030731 - (long long)h[s - 1] * pw[t - s + 1] % 20030731) %
         20030731;
}
bool check(int x, int y, int len) {
  return gethash(x, x + len - 1) == gethash(y, y + len - 1);
}
int main() {
  scanf("%d%s", &n, s + 1);
  pw[0] = 1;
  for (int i = (1); i <= (n); i++) {
    h[i] = (26ll * h[i - 1] + s[i] - 'a') % 20030731;
    pw[i] = 26ll * pw[i - 1] % 20030731;
  }
  if (n & 1) {
    len = -1;
    l = (n + 1) / 2;
    r = (n + 1) / 2;
  } else {
    len = 1;
    l = n / 2;
    r = n / 2 + 1;
  }
  for (int i = (1); i <= ((n + 1) / 2); i++) {
    (len) = (r - l + 1) < (len) ? (r - l + 1) : (len);
    while (len > 0 && !check(l, r - len + 1, len)) len -= 2;
    ans[i] = len;
    l--;
    r++;
    len += 2;
  }
  reverse(ans + 1, ans + 1 + (n + 1) / 2);
  for (int i = (1); i <= ((n + 1) / 2); i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}

### Prompt

Please create a solution in CPP 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;
char s[1000005];
int n, h[1000005], pw[1000005], ans[1000005], l, r, len;
int gethash(int s, int t) {
  return (h[t] + 20030731 - (long long)h[s - 1] * pw[t - s + 1] % 20030731) %
         20030731;
}
bool check(int x, int y, int len) {
  return gethash(x, x + len - 1) == gethash(y, y + len - 1);
}
int main() {
  scanf("%d%s", &n, s + 1);
  pw[0] = 1;
  for (int i = (1); i <= (n); i++) {
    h[i] = (26ll * h[i - 1] + s[i] - 'a') % 20030731;
    pw[i] = 26ll * pw[i - 1] % 20030731;
  }
  if (n & 1) {
    len = -1;
    l = (n + 1) / 2;
    r = (n + 1) / 2;
  } else {
    len = 1;
    l = n / 2;
    r = n / 2 + 1;
  }
  for (int i = (1); i <= ((n + 1) / 2); i++) {
    (len) = (r - l + 1) < (len) ? (r - l + 1) : (len);
    while (len > 0 && !check(l, r - len + 1, len)) len -= 2;
    ans[i] = len;
    l--;
    r++;
    len += 2;
  }
  reverse(ans + 1, ans + 1 + (n + 1) / 2);
  for (int i = (1); i <= ((n + 1) / 2); i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}

```