#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 6;
long long p = 1e9 + 7, x = 12345678, invx = 325842698;
int n, ans[MAXN];
string s;
long long px[MAXN], pix[MAXN];
long long h[MAXN];
long long ha(int i, int j) { return (h[j + 1] + p - h[i]) * pix[i] % p; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> s;
  px[0] = pix[0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    px[i] = x * px[i - 1] % p;
    pix[i] = invx * pix[i - 1] % p;
  }
  h[0] = 0;
  for (int i = 1; i <= n; ++i) {
    h[i] = (h[i - 1] + s[i - 1] * px[i - 1]) % p;
  }
  int l = (n - 1) / 2, r = n / 2, m = r - l;
  while (l >= 0) {
    int x = ans[l + 1];
    x += 2;
    x = min(x, m);
    if (x % 2 == 0) x--;
    for (int j = x; j >= -1; j -= 2) {
      if (j == -1) {
        ans[l] = -1;
      } else {
        if (ha(l, l + j - 1) == ha(r - j + 1, r)) {
          ans[l] = j;
          break;
        }
      }
    }
    l--;
    r++;
    m += 2;
  }
  for (int i = 0; i <= (n - 1) / 2; ++i) cout << ans[i] << "\n";
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 + 6;
long long p = 1e9 + 7, x = 12345678, invx = 325842698;
int n, ans[MAXN];
string s;
long long px[MAXN], pix[MAXN];
long long h[MAXN];
long long ha(int i, int j) { return (h[j + 1] + p - h[i]) * pix[i] % p; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> s;
  px[0] = pix[0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    px[i] = x * px[i - 1] % p;
    pix[i] = invx * pix[i - 1] % p;
  }
  h[0] = 0;
  for (int i = 1; i <= n; ++i) {
    h[i] = (h[i - 1] + s[i - 1] * px[i - 1]) % p;
  }
  int l = (n - 1) / 2, r = n / 2, m = r - l;
  while (l >= 0) {
    int x = ans[l + 1];
    x += 2;
    x = min(x, m);
    if (x % 2 == 0) x--;
    for (int j = x; j >= -1; j -= 2) {
      if (j == -1) {
        ans[l] = -1;
      } else {
        if (ha(l, l + j - 1) == ha(r - j + 1, r)) {
          ans[l] = j;
          break;
        }
      }
    }
    l--;
    r++;
    m += 2;
  }
  for (int i = 0; i <= (n - 1) / 2; ++i) cout << ans[i] << "\n";
}

```