#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
int base = 10;
char s[2000000];
int n;
long long has[1000015], pp[1000015];
void shash() {
  for (int i = 1; i <= n; ++i) {
    has[i] = (has[i - 1] * base + s[i]) % mod;
    pp[i] = pp[i - 1] * base % mod;
  }
}
long long get_hash(int s, int e) {
  long long res = (has[e] - pp[e - s + 1] * has[s - 1]) % mod;
  if (res < 0) res = res + mod;
  return res;
}
int ans[1000015];
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  pp[0] = 1;
  shash();
  memset(ans, -1, sizeof(ans));
  int i;
  for (i = (n + 1) / 2; i; --i) {
    int l = i, r = n + 1 - i;
    if (l == r)
      ans[i] = -1;
    else {
      for (int j = ans[i + 1] + 2; ~j; j -= 2) {
        if (get_hash(l, l + j - 1) == get_hash(r - j + 1, r)) {
          ans[i] = j;
          break;
        }
      }
    }
  }
  i = 1;
  while (i <= (n + 1) / 2) {
    cout << ans[i] << " ";
    i++;
  }
  return 0;
}

### 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 mod = 1e9 + 7;
int base = 10;
char s[2000000];
int n;
long long has[1000015], pp[1000015];
void shash() {
  for (int i = 1; i <= n; ++i) {
    has[i] = (has[i - 1] * base + s[i]) % mod;
    pp[i] = pp[i - 1] * base % mod;
  }
}
long long get_hash(int s, int e) {
  long long res = (has[e] - pp[e - s + 1] * has[s - 1]) % mod;
  if (res < 0) res = res + mod;
  return res;
}
int ans[1000015];
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  pp[0] = 1;
  shash();
  memset(ans, -1, sizeof(ans));
  int i;
  for (i = (n + 1) / 2; i; --i) {
    int l = i, r = n + 1 - i;
    if (l == r)
      ans[i] = -1;
    else {
      for (int j = ans[i + 1] + 2; ~j; j -= 2) {
        if (get_hash(l, l + j - 1) == get_hash(r - j + 1, r)) {
          ans[i] = j;
          break;
        }
      }
    }
  }
  i = 1;
  while (i <= (n + 1) / 2) {
    cout << ans[i] << " ";
    i++;
  }
  return 0;
}

```