#include <bits/stdc++.h>
using namespace std;
const int N = 1000050;
int pw[N], h[N], ans[N], n, l, r, len, cnt;
char s[N];
inline int S(int x, int l) {
  return (h[x + l - 1] - (long long)h[x - 1] * pw[l] % (998244353) +
          (998244353)) %
         (998244353);
}
int main() {
  scanf("%d%s", &n, s + 1);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i)
    pw[i] = (long long)pw[i - 1] * (31) % (998244353);
  for (int i = 1; i <= n; ++i)
    h[i] =
        ((long long)h[i - 1] * (31) % (998244353) + (s[i] - 'a')) % (998244353);
  if (n & 1)
    l = r = n / 2 + 1;
  else
    l = n / 2, r = l + 1;
  len = 1;
  for (; r <= n; len += 2) {
    for (; len != -1 && S(l, len) != S(r - len + 1, len) || len >= r - l + 1;)
      len -= 2;
    ans[++cnt] = len;
    --l;
    ++r;
  }
  for (int i = cnt; i; --i) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 = 1000050;
int pw[N], h[N], ans[N], n, l, r, len, cnt;
char s[N];
inline int S(int x, int l) {
  return (h[x + l - 1] - (long long)h[x - 1] * pw[l] % (998244353) +
          (998244353)) %
         (998244353);
}
int main() {
  scanf("%d%s", &n, s + 1);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i)
    pw[i] = (long long)pw[i - 1] * (31) % (998244353);
  for (int i = 1; i <= n; ++i)
    h[i] =
        ((long long)h[i - 1] * (31) % (998244353) + (s[i] - 'a')) % (998244353);
  if (n & 1)
    l = r = n / 2 + 1;
  else
    l = n / 2, r = l + 1;
  len = 1;
  for (; r <= n; len += 2) {
    for (; len != -1 && S(l, len) != S(r - len + 1, len) || len >= r - l + 1;)
      len -= 2;
    ans[++cnt] = len;
    --l;
    ++r;
  }
  for (int i = cnt; i; --i) printf("%d ", ans[i]);
  return 0;
}

```