#include <bits/stdc++.h>
using namespace std;
const long long bas = 1331, Mod = 1e9 + 7;
long long n, ans[1000000 + 5], has[1000000 + 5], _[1000000 + 5];
char ch[1000000 + 5];
inline long long Calc(register long long x, register long long y) {
  return (has[y] - has[x - 1] * _[y - x + 1] % Mod + Mod) % Mod;
}
signed main() {
  memset(ans, -1, sizeof(ans));
  scanf("%lld", &n);
  scanf("%s", ch + 1);
  _[0] = 1;
  for (register long long i = 1; i <= n; i++) {
    has[i] = has[i - 1] * bas % Mod + ch[i] - 96;
    _[i] = _[i - 1] * bas % Mod;
  }
  register long long L, R;
  if (!(n & 1)) {
    L = n / 2;
    R = L + 1;
    ans[n / 2] = (ch[L] == ch[R]) ? 1 : -1;
  } else
    L = R = n / 2 + 1;
  for (register long long i = (n + 1) / 2 - 1; i >= 1; i--) {
    L--;
    R++;
    for (register long long j = ans[i + 1] + 2; j >= 0; j -= 2) {
      if (Calc(L, L + j - 1) == Calc(R - j + 1, R)) {
        ans[i] = j;
        break;
      }
    }
  }
  for (register long long i = 1; i <= (n + 1) / 2; i++) printf("%lld ", ans[i]);
  return 0;
}

### Prompt

Your task is to create 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 long long bas = 1331, Mod = 1e9 + 7;
long long n, ans[1000000 + 5], has[1000000 + 5], _[1000000 + 5];
char ch[1000000 + 5];
inline long long Calc(register long long x, register long long y) {
  return (has[y] - has[x - 1] * _[y - x + 1] % Mod + Mod) % Mod;
}
signed main() {
  memset(ans, -1, sizeof(ans));
  scanf("%lld", &n);
  scanf("%s", ch + 1);
  _[0] = 1;
  for (register long long i = 1; i <= n; i++) {
    has[i] = has[i - 1] * bas % Mod + ch[i] - 96;
    _[i] = _[i - 1] * bas % Mod;
  }
  register long long L, R;
  if (!(n & 1)) {
    L = n / 2;
    R = L + 1;
    ans[n / 2] = (ch[L] == ch[R]) ? 1 : -1;
  } else
    L = R = n / 2 + 1;
  for (register long long i = (n + 1) / 2 - 1; i >= 1; i--) {
    L--;
    R++;
    for (register long long j = ans[i + 1] + 2; j >= 0; j -= 2) {
      if (Calc(L, L + j - 1) == Calc(R - j + 1, R)) {
        ans[i] = j;
        break;
      }
    }
  }
  for (register long long i = 1; i <= (n + 1) / 2; i++) printf("%lld ", ans[i]);
  return 0;
}

```