#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n;
long long tab[N];
long long hasz[N];
long long pot[N];
int dp[N];
long long get_hasz(int a, int b) {
  return ((hasz[b] - hasz[a - 1] * pot[b - a + 1]) % 1000000007 + 1000000007) %
         1000000007;
}
int main() {
  string s;
  cin >> n >> s;
  for (int i = 1; i <= n; i++) tab[i] = s[i - 1];
  pot[0] = 1;
  for (int i = 1; i <= n; i++) pot[i] = (pot[i - 1] * 997) % 1000000007;
  for (int i = 1; i <= n; i++)
    hasz[i] = (tab[i] + hasz[i - 1] * 997) % 1000000007;
  for (int i = 1; i <= (n + 1) / 2; i++) dp[i] = -1;
  for (int i = 1; i <= (n + 1) / 2; i++) {
    int j = n - i + 1;
    if (i == j) continue;
    int pocz = -1;
    int kon = 1 << 20;
    while (kon > 0) {
      int moze = pocz + kon;
      if (i - moze <= 0 || j + moze > n) {
        kon /= 2;
        continue;
      }
      long long hasz1 = get_hasz(i - moze, i + moze);
      long long hasz2 = get_hasz(j - moze, j + moze);
      if (hasz1 == hasz2) pocz = moze;
      kon /= 2;
    }
    dp[i - pocz] = max(dp[i - pocz], 2 * pocz + 1);
  }
  for (int i = 1; i <= (n + 1) / 2; i++) {
    cout << dp[i] << " ";
    dp[i + 1] = max(dp[i + 1], dp[i] - 2);
  }
}

### 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 + 10;
int n;
long long tab[N];
long long hasz[N];
long long pot[N];
int dp[N];
long long get_hasz(int a, int b) {
  return ((hasz[b] - hasz[a - 1] * pot[b - a + 1]) % 1000000007 + 1000000007) %
         1000000007;
}
int main() {
  string s;
  cin >> n >> s;
  for (int i = 1; i <= n; i++) tab[i] = s[i - 1];
  pot[0] = 1;
  for (int i = 1; i <= n; i++) pot[i] = (pot[i - 1] * 997) % 1000000007;
  for (int i = 1; i <= n; i++)
    hasz[i] = (tab[i] + hasz[i - 1] * 997) % 1000000007;
  for (int i = 1; i <= (n + 1) / 2; i++) dp[i] = -1;
  for (int i = 1; i <= (n + 1) / 2; i++) {
    int j = n - i + 1;
    if (i == j) continue;
    int pocz = -1;
    int kon = 1 << 20;
    while (kon > 0) {
      int moze = pocz + kon;
      if (i - moze <= 0 || j + moze > n) {
        kon /= 2;
        continue;
      }
      long long hasz1 = get_hasz(i - moze, i + moze);
      long long hasz2 = get_hasz(j - moze, j + moze);
      if (hasz1 == hasz2) pocz = moze;
      kon /= 2;
    }
    dp[i - pocz] = max(dp[i - pocz], 2 * pocz + 1);
  }
  for (int i = 1; i <= (n + 1) / 2; i++) {
    cout << dp[i] << " ";
    dp[i + 1] = max(dp[i + 1], dp[i] - 2);
  }
}

```