#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const int base = 311;
const int mod = 1e9 + 7;
int n, in[N];
long long H[N], Pow[N];
string a;
long long get(int i, int j) {
  return (H[j] - H[i - 1] * Pow[j - i + 1] + 1ll * mod * mod) % mod;
}
void Init() {
  Pow[0] = 1;
  for (int i = 1; i <= n; ++i) {
    Pow[i] = (Pow[i - 1] * base) % mod;
    H[i] = (H[i - 1] * base + a[i - 1]) % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  cin >> a;
  Init();
  in[(n + 3) / 2] = -1;
  for (int i = (n + 1) / 2; i > 0; --i) {
    int vt = n - i + 1;
    in[i] = -1;
    if (vt == i) continue;
    for (int j = in [i + 1] + 2; j > 0; j -= 2)
      if (get(i, i + j - 1) == get(vt - j + 1, vt)) {
        in[i] = j;
        break;
      }
  }
  for (int i = 1; i <= (n + 1) / 2; ++i) cout << in[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;
const int N = 1e6 + 5;
const int base = 311;
const int mod = 1e9 + 7;
int n, in[N];
long long H[N], Pow[N];
string a;
long long get(int i, int j) {
  return (H[j] - H[i - 1] * Pow[j - i + 1] + 1ll * mod * mod) % mod;
}
void Init() {
  Pow[0] = 1;
  for (int i = 1; i <= n; ++i) {
    Pow[i] = (Pow[i - 1] * base) % mod;
    H[i] = (H[i - 1] * base + a[i - 1]) % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  cin >> a;
  Init();
  in[(n + 3) / 2] = -1;
  for (int i = (n + 1) / 2; i > 0; --i) {
    int vt = n - i + 1;
    in[i] = -1;
    if (vt == i) continue;
    for (int j = in [i + 1] + 2; j > 0; j -= 2)
      if (get(i, i + j - 1) == get(vt - j + 1, vt)) {
        in[i] = j;
        break;
      }
  }
  for (int i = 1; i <= (n + 1) / 2; ++i) cout << in[i] << " ";
  return 0;
}

```