#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int p = 131;
const long long mod = 1e9 + 7;
long long powp[N], _hash[N];
char s[N];
int ans[N];
int n;
bool check(int x, int len) {
  long long hash1, hash2;
  hash1 = (_hash[x + len - 1] - _hash[x - 1] * powp[len] % mod + mod) % mod;
  hash2 =
      (_hash[n - x + 1] - _hash[n - x - len + 1] * powp[len] % mod + mod) % mod;
  return hash1 == hash2;
}
int main() {
  int half;
  scanf("%d", &n);
  scanf("%s", s + 1);
  powp[0] = 1;
  _hash[0] = 0;
  for (int i = 1; i <= n; i++) {
    powp[i] = powp[i - 1] * p % mod;
    _hash[i] = (_hash[i - 1] * p % mod + s[i] - 'a' + 1) % mod;
  }
  half = (n + 1) / 2;
  if (n & 1)
    ans[half] = -1;
  else if (s[half] == s[half + 1])
    ans[half] = 1;
  else
    ans[half] = -1;
  for (int i = half - 1; i; i--) {
    ans[i] = ans[i + 1] + 2;
    while (~ans[i] && !check(i, ans[i])) ans[i] -= 2;
  }
  for (int i = 1; i <= half; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 1e6 + 10;
const int p = 131;
const long long mod = 1e9 + 7;
long long powp[N], _hash[N];
char s[N];
int ans[N];
int n;
bool check(int x, int len) {
  long long hash1, hash2;
  hash1 = (_hash[x + len - 1] - _hash[x - 1] * powp[len] % mod + mod) % mod;
  hash2 =
      (_hash[n - x + 1] - _hash[n - x - len + 1] * powp[len] % mod + mod) % mod;
  return hash1 == hash2;
}
int main() {
  int half;
  scanf("%d", &n);
  scanf("%s", s + 1);
  powp[0] = 1;
  _hash[0] = 0;
  for (int i = 1; i <= n; i++) {
    powp[i] = powp[i - 1] * p % mod;
    _hash[i] = (_hash[i - 1] * p % mod + s[i] - 'a' + 1) % mod;
  }
  half = (n + 1) / 2;
  if (n & 1)
    ans[half] = -1;
  else if (s[half] == s[half + 1])
    ans[half] = 1;
  else
    ans[half] = -1;
  for (int i = half - 1; i; i--) {
    ans[i] = ans[i + 1] + 2;
    while (~ans[i] && !check(i, ans[i])) ans[i] -= 2;
  }
  for (int i = 1; i <= half; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

```