#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7;
const int P = 107;
const long long mod = 1e9 + 7;
int n;
char s[N];
int ans[N];
int cnt, l, r;
long long x, y;
long long p[N], sum[N];
inline long long mo(long long x) {
  while (x >= mod) x -= mod;
  return x;
}
inline void Hash() {
  p[0] = 1;
  for (int i = 1; i <= n; i++) {
    p[i] = p[i - 1] * P % mod;
    sum[i] = mo(sum[i - 1] * P % mod + s[i] - 'a' + 1);
  }
}
inline long long hs(int l, int r) {
  return mo(sum[r] - sum[l - 1] * p[r - l + 1] % mod + mod);
}
int main() {
  scanf("%d%s", &n, s + 1);
  Hash();
  cnt = n + 1 >> 1;
  if (n & 1)
    ans[cnt] = -1, l = r = cnt;
  else {
    l = n >> 1, r = l + 1;
    if (s[l] == s[r])
      ans[cnt] = 1;
    else
      ans[cnt] = -1;
  }
  for (int i = cnt - 1; i >= 1; i--) {
    l--, r++;
    for (int j = ans[i + 1] + 2; j >= -1; j -= 2) {
      x = hs(l, l + j - 1), y = hs(r - j + 1, r);
      if (x == y || j == -1) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= cnt; i++) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 + 7;
const int P = 107;
const long long mod = 1e9 + 7;
int n;
char s[N];
int ans[N];
int cnt, l, r;
long long x, y;
long long p[N], sum[N];
inline long long mo(long long x) {
  while (x >= mod) x -= mod;
  return x;
}
inline void Hash() {
  p[0] = 1;
  for (int i = 1; i <= n; i++) {
    p[i] = p[i - 1] * P % mod;
    sum[i] = mo(sum[i - 1] * P % mod + s[i] - 'a' + 1);
  }
}
inline long long hs(int l, int r) {
  return mo(sum[r] - sum[l - 1] * p[r - l + 1] % mod + mod);
}
int main() {
  scanf("%d%s", &n, s + 1);
  Hash();
  cnt = n + 1 >> 1;
  if (n & 1)
    ans[cnt] = -1, l = r = cnt;
  else {
    l = n >> 1, r = l + 1;
    if (s[l] == s[r])
      ans[cnt] = 1;
    else
      ans[cnt] = -1;
  }
  for (int i = cnt - 1; i >= 1; i--) {
    l--, r++;
    for (int j = ans[i + 1] + 2; j >= -1; j -= 2) {
      x = hs(l, l + j - 1), y = hs(r - j + 1, r);
      if (x == y || j == -1) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= cnt; i++) printf("%d ", ans[i]);
  return 0;
}

```