#include <bits/stdc++.h>
using namespace std;
const long long p = 151;
const long long mod = 91815541;
const int N = 1000005;
int n, mid;
char text[N];
long long Hash[N];
long long P[N];
int ans[N];
void init_hash() {
  P[0] = 1;
  Hash[0] = text[0] - 'a';
  for (int i = 1; i < n; i++) {
    Hash[i] = (Hash[i - 1] * p) % mod + text[i] - 'a';
    P[i] = (P[i - 1] * p) % mod;
  }
}
long long get_hash(int i, int j) {
  if (i == 0)
    return Hash[j];
  else
    return (Hash[j] - (Hash[i - 1] * P[j - i + 1]) % mod + mod) % mod;
}
int main() {
  cin >> n;
  int l, r;
  cin >> text;
  mid = (n - 1) >> 1;
  if (n % 2)
    l = r = mid, ans[mid] = -1;
  else {
    l = mid;
    r = mid + 1;
    if (text[l] == text[r])
      ans[mid] = 1;
    else
      ans[mid] = -1;
  }
  init_hash();
  for (int i = mid - 1; i >= 0; i--) {
    l--, r++;
    for (int j = ans[i + 1] + 2; j >= -1; j -= 2) {
      if (j == -1 || get_hash(l, l + j - 1) == get_hash(r - j + 1, r)) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 0; i <= mid; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}

### Prompt

Develop a solution in Cpp 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 long long p = 151;
const long long mod = 91815541;
const int N = 1000005;
int n, mid;
char text[N];
long long Hash[N];
long long P[N];
int ans[N];
void init_hash() {
  P[0] = 1;
  Hash[0] = text[0] - 'a';
  for (int i = 1; i < n; i++) {
    Hash[i] = (Hash[i - 1] * p) % mod + text[i] - 'a';
    P[i] = (P[i - 1] * p) % mod;
  }
}
long long get_hash(int i, int j) {
  if (i == 0)
    return Hash[j];
  else
    return (Hash[j] - (Hash[i - 1] * P[j - i + 1]) % mod + mod) % mod;
}
int main() {
  cin >> n;
  int l, r;
  cin >> text;
  mid = (n - 1) >> 1;
  if (n % 2)
    l = r = mid, ans[mid] = -1;
  else {
    l = mid;
    r = mid + 1;
    if (text[l] == text[r])
      ans[mid] = 1;
    else
      ans[mid] = -1;
  }
  init_hash();
  for (int i = mid - 1; i >= 0; i--) {
    l--, r++;
    for (int j = ans[i + 1] + 2; j >= -1; j -= 2) {
      if (j == -1 || get_hash(l, l + j - 1) == get_hash(r - j + 1, r)) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 0; i <= mid; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}

```