#include <bits/stdc++.h>
using namespace std;
char s[1000010];
int fv[1000010][2];
int hv[1000010][2];
int ans[1000010];
int n;
void pre() {
  fv[0][0] = fv[0][1] = 1;
  for (int i = 1; i <= n; i++) {
    fv[i][0] = 1LL * fv[i - 1][0] * 37 % 1000000007;
    hv[i][0] = (1LL * hv[i - 1][0] * 37 + s[i] - 'a' + 1) % 1000000007;
    fv[i][1] = 1LL * fv[i - 1][1] * 31 % 998244353;
    hv[i][1] = (1LL * hv[i - 1][1] * 31 + s[i] - 'a' + 1) % 998244353;
  }
}
pair<int, int> calc(int L, int R) {
  return pair<int, int>(
      ((hv[R][0] - 1LL * hv[L - 1][0] * fv[R - L + 1][0]) % 1000000007 +
       1000000007) %
          1000000007,
      ((hv[R][1] - 1LL * hv[L - 1][1] * fv[R - L + 1][1]) % 998244353 +
       998244353) %
          998244353);
}
int main() {
  scanf("%d%s", &n, s + 1);
  pre();
  int st = -1;
  for (int i = (n + 1) / 2; i >= 1; i--) {
    st += 2;
    st = min(st, n - 2 * i + 1);
    if (st % 2 == 0) st--;
    while (st > 0) {
      if (calc(i, i + st - 1) == calc(n + 1 - i - st + 1, n + 1 - i)) break;
      st -= 2;
    }
    if (st > 0)
      ans[i] = st;
    else
      ans[i] = -1;
  }
  for (int i = 1; i <= (n + 1) / 2; i++) printf("%d ", ans[i]);
}

### Prompt

Your challenge is to write 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;
char s[1000010];
int fv[1000010][2];
int hv[1000010][2];
int ans[1000010];
int n;
void pre() {
  fv[0][0] = fv[0][1] = 1;
  for (int i = 1; i <= n; i++) {
    fv[i][0] = 1LL * fv[i - 1][0] * 37 % 1000000007;
    hv[i][0] = (1LL * hv[i - 1][0] * 37 + s[i] - 'a' + 1) % 1000000007;
    fv[i][1] = 1LL * fv[i - 1][1] * 31 % 998244353;
    hv[i][1] = (1LL * hv[i - 1][1] * 31 + s[i] - 'a' + 1) % 998244353;
  }
}
pair<int, int> calc(int L, int R) {
  return pair<int, int>(
      ((hv[R][0] - 1LL * hv[L - 1][0] * fv[R - L + 1][0]) % 1000000007 +
       1000000007) %
          1000000007,
      ((hv[R][1] - 1LL * hv[L - 1][1] * fv[R - L + 1][1]) % 998244353 +
       998244353) %
          998244353);
}
int main() {
  scanf("%d%s", &n, s + 1);
  pre();
  int st = -1;
  for (int i = (n + 1) / 2; i >= 1; i--) {
    st += 2;
    st = min(st, n - 2 * i + 1);
    if (st % 2 == 0) st--;
    while (st > 0) {
      if (calc(i, i + st - 1) == calc(n + 1 - i - st + 1, n + 1 - i)) break;
      st -= 2;
    }
    if (st > 0)
      ans[i] = st;
    else
      ans[i] = -1;
  }
  for (int i = 1; i <= (n + 1) / 2; i++) printf("%d ", ans[i]);
}

```