#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
using ull = long long;
const ull mod = 1e9 + 7;
const ull base = 19260817;
char s[maxn];
ull prx[maxn], pw[maxn];
int n;
int ans[maxn];
ull get(int l, int r) {
  return ((prx[r] - prx[l - 1] * pw[r - l + 1]) % mod + mod) % mod;
}
int main() {
  cin >> n;
  cin >> s + 1;
  prx[0] = 0, pw[0] = 1;
  for (int i = 1; i <= n; i++) {
    prx[i] = (prx[i - 1] * base + s[i]) % mod;
    pw[i] = pw[i - 1] * base % mod;
  }
  memset(ans, -1, sizeof ans);
  int l, r;
  if (n & 1) {
    l = r = (n + 1) / 2;
    ans[l] = -1;
  } else {
    l = n / 2, r = n / 2 + 1;
    if (s[l] == s[r])
      ans[l] = 1;
    else
      ans[l] = -1;
  }
  for (l--, r++; l >= 1; l--, r++) {
    for (int d = min(r - l, ans[l + 1] + 2); d >= 1; d -= 2)
      if (get(l, l + d - 1) == get(r - d + 1, r)) {
        ans[l] = d;
        break;
      }
  }
  for (int i = 1; i <= (n + 1) / 2; i++) cout << ans[i] << " ";
  cout << endl;
}

### 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 maxn = 1e6 + 5;
using ull = long long;
const ull mod = 1e9 + 7;
const ull base = 19260817;
char s[maxn];
ull prx[maxn], pw[maxn];
int n;
int ans[maxn];
ull get(int l, int r) {
  return ((prx[r] - prx[l - 1] * pw[r - l + 1]) % mod + mod) % mod;
}
int main() {
  cin >> n;
  cin >> s + 1;
  prx[0] = 0, pw[0] = 1;
  for (int i = 1; i <= n; i++) {
    prx[i] = (prx[i - 1] * base + s[i]) % mod;
    pw[i] = pw[i - 1] * base % mod;
  }
  memset(ans, -1, sizeof ans);
  int l, r;
  if (n & 1) {
    l = r = (n + 1) / 2;
    ans[l] = -1;
  } else {
    l = n / 2, r = n / 2 + 1;
    if (s[l] == s[r])
      ans[l] = 1;
    else
      ans[l] = -1;
  }
  for (l--, r++; l >= 1; l--, r++) {
    for (int d = min(r - l, ans[l + 1] + 2); d >= 1; d -= 2)
      if (get(l, l + d - 1) == get(r - d + 1, r)) {
        ans[l] = d;
        break;
      }
  }
  for (int i = 1; i <= (n + 1) / 2; i++) cout << ans[i] << " ";
  cout << endl;
}

```