#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
const int P = 1e9 + 7;
int n;
int ans[N];
char s[N];
int pw[N], hsh[N];
int subHash(int lst, int len);
int main() {
  scanf("%d%s", &n, s + 1);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) {
    pw[i] = pw[i - 1] * 26LL % P;
    hsh[i] = (hsh[i - 1] * 26LL + (s[i] - 'a')) % P;
  }
  int cur = -1;
  for (int i = (n + 1) / 2; i; --i) {
    cur += 2;
    if (i * 2 == n + 1) cur = -1;
    while (cur != -1 && subHash(i + cur - 1, cur) != subHash(n - i + 1, cur))
      cur -= 2;
    ans[i] = cur;
  }
  for (int i = 1; i <= (n + 1) / 2; ++i) printf("%d ", ans[i]);
  return 0;
}
int subHash(int lst, int len) {
  int ret = (hsh[lst] - hsh[lst - len] * (long long)pw[len]) % P;
  if (ret < 0) ret += P;
  return ret;
}

### Prompt

Please formulate 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 = 1000010;
const int P = 1e9 + 7;
int n;
int ans[N];
char s[N];
int pw[N], hsh[N];
int subHash(int lst, int len);
int main() {
  scanf("%d%s", &n, s + 1);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) {
    pw[i] = pw[i - 1] * 26LL % P;
    hsh[i] = (hsh[i - 1] * 26LL + (s[i] - 'a')) % P;
  }
  int cur = -1;
  for (int i = (n + 1) / 2; i; --i) {
    cur += 2;
    if (i * 2 == n + 1) cur = -1;
    while (cur != -1 && subHash(i + cur - 1, cur) != subHash(n - i + 1, cur))
      cur -= 2;
    ans[i] = cur;
  }
  for (int i = 1; i <= (n + 1) / 2; ++i) printf("%d ", ans[i]);
  return 0;
}
int subHash(int lst, int len) {
  int ret = (hsh[lst] - hsh[lst - len] * (long long)pw[len]) % P;
  if (ret < 0) ret += P;
  return ret;
}

```