#include <bits/stdc++.h>
const int N = 1e6 + 5;
const long long BASE = 127;
const long long MOD = 998244353;
char s[N];
long long HASH[N], HFac[N];
int ans[N];
int n, cnt = 0, l, r, len = 1;
long long get_hash(int l, int len) {
  return (HASH[l + len - 1] + MOD - (HASH[l - 1] * HFac[len]) % MOD) % MOD;
}
void pre() {
  HFac[0] = 1;
  for (int i = 1; i <= n; i++) HFac[i] = (HFac[i - 1] * BASE) % MOD;
  HASH[0] = 0;
  for (int i = 1; i <= n; i++)
    HASH[i] = ((HASH[i - 1] * BASE) % MOD + s[i] - 'a') % MOD;
}
int main() {
  scanf("%d %s", &n, s + 1);
  pre();
  if (n & 1)
    l = r = n / 2 + 1;
  else
    l = n / 2, r = l + 1;
  while (r <= n && l >= 1) {
    while ((len != -1 && get_hash(l, len) != get_hash(r - len + 1, len)) ||
           (len >= r - l + 1))
      len -= 2;
    ans[++cnt] = len;
    l--;
    r++;
    len += 2;
  }
  for (int i = cnt; i; i--) printf("%d%c", ans[i], n == i ? '\n' : ' ');
  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>
const int N = 1e6 + 5;
const long long BASE = 127;
const long long MOD = 998244353;
char s[N];
long long HASH[N], HFac[N];
int ans[N];
int n, cnt = 0, l, r, len = 1;
long long get_hash(int l, int len) {
  return (HASH[l + len - 1] + MOD - (HASH[l - 1] * HFac[len]) % MOD) % MOD;
}
void pre() {
  HFac[0] = 1;
  for (int i = 1; i <= n; i++) HFac[i] = (HFac[i - 1] * BASE) % MOD;
  HASH[0] = 0;
  for (int i = 1; i <= n; i++)
    HASH[i] = ((HASH[i - 1] * BASE) % MOD + s[i] - 'a') % MOD;
}
int main() {
  scanf("%d %s", &n, s + 1);
  pre();
  if (n & 1)
    l = r = n / 2 + 1;
  else
    l = n / 2, r = l + 1;
  while (r <= n && l >= 1) {
    while ((len != -1 && get_hash(l, len) != get_hash(r - len + 1, len)) ||
           (len >= r - l + 1))
      len -= 2;
    ans[++cnt] = len;
    l--;
    r++;
    len += 2;
  }
  for (int i = cnt; i; i--) printf("%d%c", ans[i], n == i ? '\n' : ' ');
  return 0;
}

```