#include <bits/stdc++.h>
using namespace std;
vector<long long> ha(1000001), t(1000001);
vector<int> ans(500001);
long long gethash(int l, int r) {
  long long hh;
  if (l == 0) return ha[r];
  hh = (ha[r] - (ha[l - 1] * t[r - l + 1]) % 1000000007 + 1000000007) %
       1000000007;
  return hh;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  string a;
  cin >> a;
  ha[0] = a[0], t[0] = 1;
  for (int i = 1; i < n; i++) {
    ha[i] = (ha[i - 1] * 29 + a[i]) % 1000000007;
    t[i] = t[i - 1] * 29 % 1000000007;
  }
  int l, r;
  if (n % 2 == 1) {
    ans[(n - 1) / 2] = -1;
    l = r = (n - 1) / 2;
    for (int i = (n - 1) / 2 - 1; i >= 0; i--) {
      l--;
      r++;
      int j;
      for (j = ans[i + 1] + 2; j >= 1; j -= 2) {
        if (gethash(l, l + j - 1) == gethash(r - j + 1, r)) {
          ans[i] = j;
          break;
        }
      }
      if (j < 1) {
        ans[i] = -1;
      }
    }
  } else {
    if (a[n / 2] == a[n / 2 - 1]) {
      ans[n / 2 - 1] = 1;
    } else {
      ans[n / 2 - 1] = -1;
    }
    l = n / 2 - 1;
    r = n / 2;
    for (int i = n / 2 - 2; i >= 0; i--) {
      l--;
      r++;
      int j;
      for (j = ans[i + 1] + 2; j >= 1; j -= 2) {
        if (gethash(l, l + j - 1) == gethash(r - j + 1, r)) {
          ans[i] = j;
          break;
        }
      }
      if (j < 1) {
        ans[i] = -1;
      }
    }
  }
  for (int i = 0; i < (n + 1) / 2 - 1; i++) {
    printf("%d ", ans[i]);
  }
  printf("%d", ans[(n + 1) / 2 - 1]);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
vector<long long> ha(1000001), t(1000001);
vector<int> ans(500001);
long long gethash(int l, int r) {
  long long hh;
  if (l == 0) return ha[r];
  hh = (ha[r] - (ha[l - 1] * t[r - l + 1]) % 1000000007 + 1000000007) %
       1000000007;
  return hh;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  string a;
  cin >> a;
  ha[0] = a[0], t[0] = 1;
  for (int i = 1; i < n; i++) {
    ha[i] = (ha[i - 1] * 29 + a[i]) % 1000000007;
    t[i] = t[i - 1] * 29 % 1000000007;
  }
  int l, r;
  if (n % 2 == 1) {
    ans[(n - 1) / 2] = -1;
    l = r = (n - 1) / 2;
    for (int i = (n - 1) / 2 - 1; i >= 0; i--) {
      l--;
      r++;
      int j;
      for (j = ans[i + 1] + 2; j >= 1; j -= 2) {
        if (gethash(l, l + j - 1) == gethash(r - j + 1, r)) {
          ans[i] = j;
          break;
        }
      }
      if (j < 1) {
        ans[i] = -1;
      }
    }
  } else {
    if (a[n / 2] == a[n / 2 - 1]) {
      ans[n / 2 - 1] = 1;
    } else {
      ans[n / 2 - 1] = -1;
    }
    l = n / 2 - 1;
    r = n / 2;
    for (int i = n / 2 - 2; i >= 0; i--) {
      l--;
      r++;
      int j;
      for (j = ans[i + 1] + 2; j >= 1; j -= 2) {
        if (gethash(l, l + j - 1) == gethash(r - j + 1, r)) {
          ans[i] = j;
          break;
        }
      }
      if (j < 1) {
        ans[i] = -1;
      }
    }
  }
  for (int i = 0; i < (n + 1) / 2 - 1; i++) {
    printf("%d ", ans[i]);
  }
  printf("%d", ans[(n + 1) / 2 - 1]);
  return 0;
}

```