#include <bits/stdc++.h>
using namespace std;
char ch[1000005];
int n, f[1000005];
long long h[1000005], K = 119, cf[1000005];
bool pd(int p, int q, int le) {
  int l = p - le + 1, r = p + le - 1, x = q - le + 1, y = q + le - 1;
  return ((h[r] - h[l - 1] * cf[2 * le - 1] % 1000000009 + 1000000009) %
              1000000009 ==
          (h[y] - h[x - 1] * cf[2 * le - 1] % 1000000009 + 1000000009) %
              1000000009);
}
int main() {
  cin >> n;
  scanf("\n");
  cf[0] = 1;
  for (int i = 1; i <= n; i++) {
    ch[i] = getchar();
    h[i] = (h[i - 1] * K % 1000000009 + ch[i] - 'a') % 1000000009;
    cf[i] = cf[i - 1] * K % 1000000009;
  }
  int n1 = (n + 1) >> 1;
  for (int i = 1; i <= n1; i++) {
    if (n - i + 1 == i) break;
    int l = 1, r = i;
    while (l + 1 < r) {
      int mid = (l + r) >> 1;
      if (pd(i, n - i + 1, mid))
        l = mid;
      else
        r = mid;
    }
    if (pd(i, n - i + 1, r))
      f[i - r + 1] = max(f[i - r + 1], 2 * r - 1);
    else if (pd(i, n - i + 1, l))
      f[i - l + 1] = max(f[i - l + 1], 2 * l - 1);
  }
  for (int i = 1; i <= n1; i++) {
    f[i] = max(f[i], f[i - 1] - 2);
    if (f[i] == 0)
      printf("-1 ");
    else
      printf("%d ", f[i]);
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 ch[1000005];
int n, f[1000005];
long long h[1000005], K = 119, cf[1000005];
bool pd(int p, int q, int le) {
  int l = p - le + 1, r = p + le - 1, x = q - le + 1, y = q + le - 1;
  return ((h[r] - h[l - 1] * cf[2 * le - 1] % 1000000009 + 1000000009) %
              1000000009 ==
          (h[y] - h[x - 1] * cf[2 * le - 1] % 1000000009 + 1000000009) %
              1000000009);
}
int main() {
  cin >> n;
  scanf("\n");
  cf[0] = 1;
  for (int i = 1; i <= n; i++) {
    ch[i] = getchar();
    h[i] = (h[i - 1] * K % 1000000009 + ch[i] - 'a') % 1000000009;
    cf[i] = cf[i - 1] * K % 1000000009;
  }
  int n1 = (n + 1) >> 1;
  for (int i = 1; i <= n1; i++) {
    if (n - i + 1 == i) break;
    int l = 1, r = i;
    while (l + 1 < r) {
      int mid = (l + r) >> 1;
      if (pd(i, n - i + 1, mid))
        l = mid;
      else
        r = mid;
    }
    if (pd(i, n - i + 1, r))
      f[i - r + 1] = max(f[i - r + 1], 2 * r - 1);
    else if (pd(i, n - i + 1, l))
      f[i - l + 1] = max(f[i - l + 1], 2 * l - 1);
  }
  for (int i = 1; i <= n1; i++) {
    f[i] = max(f[i], f[i - 1] - 2);
    if (f[i] == 0)
      printf("-1 ");
    else
      printf("%d ", f[i]);
  }
}

```