#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 11;
int n, ans[N], t;
char s[N];
int hsh[2][N], p[2][N];
const int mod[] = {19260817, 988244353};
inline int getval(int id, int l, int r) {
  return (hsh[id][r] - 1ll * hsh[id][l - 1] * p[id][r - l + 1] % mod[id] +
          mod[id]) %
         mod[id];
}
int main() {
  scanf("%d\n", &n);
  gets(s + 1);
  p[0][0] = p[1][0] = 1;
  for (register int i = 1; i <= n; ++i)
    for (register int j = 0; j <= 1; ++j) {
      p[j][i] = 1ll * p[j][i - 1] * 97 % mod[j];
      hsh[j][i] = (hsh[j][i - 1] * 97ll + s[i] - 'a' + 1) % mod[j];
    }
  for (register int i = (n + 1) / 2; i >= 1; --i) {
    if (i == (n + 1) / 2) {
      if (n & 1)
        ans[i] = -1;
      else {
        if (s[(n + 1) / 2] == s[(n + 1) / 2 + 1])
          ans[i] = 1;
        else
          ans[i] = -1;
      }
    } else {
      t = ans[i + 1] + 2;
      while (t != -1 &&
             (getval(0, i, i + t - 1) != getval(0, n - i - t + 2, n - i + 1) ||
              (getval(1, i, i + t - 1) != getval(1, n - i - t + 2, n - i + 1))))
        t -= 2;
      ans[i] = t;
    }
  }
  for (register int i = 1; i <= (n + 1) / 2; ++i) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Please formulate 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 + 11;
int n, ans[N], t;
char s[N];
int hsh[2][N], p[2][N];
const int mod[] = {19260817, 988244353};
inline int getval(int id, int l, int r) {
  return (hsh[id][r] - 1ll * hsh[id][l - 1] * p[id][r - l + 1] % mod[id] +
          mod[id]) %
         mod[id];
}
int main() {
  scanf("%d\n", &n);
  gets(s + 1);
  p[0][0] = p[1][0] = 1;
  for (register int i = 1; i <= n; ++i)
    for (register int j = 0; j <= 1; ++j) {
      p[j][i] = 1ll * p[j][i - 1] * 97 % mod[j];
      hsh[j][i] = (hsh[j][i - 1] * 97ll + s[i] - 'a' + 1) % mod[j];
    }
  for (register int i = (n + 1) / 2; i >= 1; --i) {
    if (i == (n + 1) / 2) {
      if (n & 1)
        ans[i] = -1;
      else {
        if (s[(n + 1) / 2] == s[(n + 1) / 2 + 1])
          ans[i] = 1;
        else
          ans[i] = -1;
      }
    } else {
      t = ans[i + 1] + 2;
      while (t != -1 &&
             (getval(0, i, i + t - 1) != getval(0, n - i - t + 2, n - i + 1) ||
              (getval(1, i, i + t - 1) != getval(1, n - i - t + 2, n - i + 1))))
        t -= 2;
      ans[i] = t;
    }
  }
  for (register int i = 1; i <= (n + 1) / 2; ++i) printf("%d ", ans[i]);
  return 0;
}

```