#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e6 + 5;
const int MOD = 1e9 + 7;
const int B = 31;
int f[N];
char s[N];
ll pw[N], hs[N];
int get_hs(int i, int j) {
  return ((hs[j] - hs[i - 1] * pw[j - i + 1] % MOD) + MOD) % MOD;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int n;
  cin >> n >> (s + 1);
  pw[0] = 1;
  hs[0] = 0;
  for (int i = 1; i <= n; i++) {
    hs[i] = (hs[i - 1] * B + s[i] - 'a' + 1) % MOD;
    pw[i] = pw[i - 1] * B % MOD;
  }
  for (int i = (n + 1) / 2; i > 0; i--) {
    f[i] = min(n + 2 - 2 * i - 1, f[i + 1] + 2);
    if (!(f[i] & 1)) f[i]--;
    while (~f[i] &&
           get_hs(i, i + f[i] - 1) != get_hs(n - i - f[i] + 2, n - i + 1))
      f[i] -= 2;
  }
  for (int i = 1; i <= (n + 1) / 2; i++) cout << f[i] << ' ';
}

### Prompt

Create a solution in Cpp for 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;
using ll = long long;
const int N = 1e6 + 5;
const int MOD = 1e9 + 7;
const int B = 31;
int f[N];
char s[N];
ll pw[N], hs[N];
int get_hs(int i, int j) {
  return ((hs[j] - hs[i - 1] * pw[j - i + 1] % MOD) + MOD) % MOD;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int n;
  cin >> n >> (s + 1);
  pw[0] = 1;
  hs[0] = 0;
  for (int i = 1; i <= n; i++) {
    hs[i] = (hs[i - 1] * B + s[i] - 'a' + 1) % MOD;
    pw[i] = pw[i - 1] * B % MOD;
  }
  for (int i = (n + 1) / 2; i > 0; i--) {
    f[i] = min(n + 2 - 2 * i - 1, f[i + 1] + 2);
    if (!(f[i] & 1)) f[i]--;
    while (~f[i] &&
           get_hs(i, i + f[i] - 1) != get_hs(n - i - f[i] + 2, n - i + 1))
      f[i] -= 2;
  }
  for (int i = 1; i <= (n + 1) / 2; i++) cout << f[i] << ' ';
}

```