#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e6 + 100;
int n;
char s[N];
int z[N];
pair<int, int> st[N];
int m;
int ans[N];
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  for (int i = n - 1; i > 0; i--) s[2 * i] = s[i];
  for (int i = 1; i < 2 * n; i += 2) s[i] = s[2 * n - 1 - i];
  int l = 0, r = 0;
  for (int i = 1; i < n; i++) {
    if (i <= r) z[i] = min(r - i + 1, z[l + r - i + 1]);
    while (i > z[i] && s[i - z[i] - 1] == s[i + z[i]]) z[i]++;
    if (i + z[i] - 1 > r) {
      l = i - z[i];
      r = i + z[i] - 1;
    }
  }
  for (int i = 1; i < n; i += 2) {
    l = i - z[i];
    while (m > 0 && st[m - 1].first >= l) m--;
    st[m++] = make_pair(l, i);
  }
  for (int i = 0; i <= n; i++) ans[i] = -1;
  for (int i = 0; i < m; i++) {
    r = st[i].second;
    if (i + 1 < m) r = min(r, st[i + 1].first);
    for (int j = st[i].first; j < r; j++) ans[j] = st[i].second - j;
  }
  for (int i = 0; i < n; i += 2) printf("%d ", ans[i]);
  printf("\n");
  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 = (int)2e6 + 100;
int n;
char s[N];
int z[N];
pair<int, int> st[N];
int m;
int ans[N];
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  for (int i = n - 1; i > 0; i--) s[2 * i] = s[i];
  for (int i = 1; i < 2 * n; i += 2) s[i] = s[2 * n - 1 - i];
  int l = 0, r = 0;
  for (int i = 1; i < n; i++) {
    if (i <= r) z[i] = min(r - i + 1, z[l + r - i + 1]);
    while (i > z[i] && s[i - z[i] - 1] == s[i + z[i]]) z[i]++;
    if (i + z[i] - 1 > r) {
      l = i - z[i];
      r = i + z[i] - 1;
    }
  }
  for (int i = 1; i < n; i += 2) {
    l = i - z[i];
    while (m > 0 && st[m - 1].first >= l) m--;
    st[m++] = make_pair(l, i);
  }
  for (int i = 0; i <= n; i++) ans[i] = -1;
  for (int i = 0; i < m; i++) {
    r = st[i].second;
    if (i + 1 < m) r = min(r, st[i + 1].first);
    for (int j = st[i].first; j < r; j++) ans[j] = st[i].second - j;
  }
  for (int i = 0; i < n; i += 2) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

```