#include <bits/stdc++.h>
using namespace std;
const int N = 4e6 + 10;
void cmax(int& x, int y) {
  if (x < y) {
    x = y;
  }
}
int n, m, f[N], tag[N];
char s[N], t[N];
int main() {
  scanf("%d%s", &n, s);
  memset(tag, -1, sizeof tag);
  for (int i = 0; i < n; ++i) {
    t[m++] = s[i];
    t[m++] = s[n - i - 1];
  }
  for (int i = 0; i < m; ++i) {
    s[i << 1 | 1] = t[i];
  }
  for (int i = 0; i <= m << 1; i += 2) {
    s[i] = '$';
  }
  for (int i = 1, j = 0, k = 0; i < m; ++i) {
    f[i] = k > i ? min(k - i, f[2 * j - i]) : 0;
    while (i - f[i] >= 0 && i + f[i] <= (m << 1) &&
           s[i + f[i]] == s[i - f[i]]) {
      ++f[i];
    }
    if (i + f[i] > k) {
      j = i;
      k = i + f[i];
    }
    if (i % 4 == 2) {
      if ((f[i] - 1 >> 1) & 1) {
        cmax(tag[i - f[i] + 2], f[i] - 1 >> 1);
      } else {
        cmax(tag[i - f[i] + 4], (f[i] - 1 >> 1) - 1);
      }
    }
  }
  int answer = -1;
  for (int i = 1, j = 0; i < m; i += 4, j += 2) {
    cmax(answer, tag[i] + j);
    printf("%d%c", answer - j, " \n"[i + 4 > (m << 1)]);
  }
  return 0;
}

### 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;
const int N = 4e6 + 10;
void cmax(int& x, int y) {
  if (x < y) {
    x = y;
  }
}
int n, m, f[N], tag[N];
char s[N], t[N];
int main() {
  scanf("%d%s", &n, s);
  memset(tag, -1, sizeof tag);
  for (int i = 0; i < n; ++i) {
    t[m++] = s[i];
    t[m++] = s[n - i - 1];
  }
  for (int i = 0; i < m; ++i) {
    s[i << 1 | 1] = t[i];
  }
  for (int i = 0; i <= m << 1; i += 2) {
    s[i] = '$';
  }
  for (int i = 1, j = 0, k = 0; i < m; ++i) {
    f[i] = k > i ? min(k - i, f[2 * j - i]) : 0;
    while (i - f[i] >= 0 && i + f[i] <= (m << 1) &&
           s[i + f[i]] == s[i - f[i]]) {
      ++f[i];
    }
    if (i + f[i] > k) {
      j = i;
      k = i + f[i];
    }
    if (i % 4 == 2) {
      if ((f[i] - 1 >> 1) & 1) {
        cmax(tag[i - f[i] + 2], f[i] - 1 >> 1);
      } else {
        cmax(tag[i - f[i] + 4], (f[i] - 1 >> 1) - 1);
      }
    }
  }
  int answer = -1;
  for (int i = 1, j = 0; i < m; i += 4, j += 2) {
    cmax(answer, tag[i] + j);
    printf("%d%c", answer - j, " \n"[i + 4 > (m << 1)]);
  }
  return 0;
}

```