#include <bits/stdc++.h>
using namespace std;
const long long P = 1e9 + 7;
const int maxn = 2005000;
char s[maxn];
int n, i, c;
int ans[maxn];
long long h[maxn], v[maxn], fac[maxn] = {1};
long long get(int l, int r) {
  return (h[r] - h[l - 1] * fac[r - l + 1] % P + P) % P;
}
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) {
    h[i] = (h[i - 1] * 419 + s[i] - 'a') % P;
    fac[i] = fac[i - 1] * 419 % P;
  }
  memset(ans, -1, sizeof(ans));
  for (int i = (n + 1) / 2; i; i--) {
    int j = n - i + 1;
    if (i == j) {
      ans[i] = -1;
      continue;
    }
    ans[i] = ans[i + 1] + 2;
    while (~ans[i]) {
      if (get(i, i + ans[i] - 1) == get(j - ans[i] + 1, j)) break;
      ans[i] -= 2;
    }
  }
  for (int i = 1; i <= (n + 1) / 2; i++) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Your task is to create 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 long long P = 1e9 + 7;
const int maxn = 2005000;
char s[maxn];
int n, i, c;
int ans[maxn];
long long h[maxn], v[maxn], fac[maxn] = {1};
long long get(int l, int r) {
  return (h[r] - h[l - 1] * fac[r - l + 1] % P + P) % P;
}
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) {
    h[i] = (h[i - 1] * 419 + s[i] - 'a') % P;
    fac[i] = fac[i - 1] * 419 % P;
  }
  memset(ans, -1, sizeof(ans));
  for (int i = (n + 1) / 2; i; i--) {
    int j = n - i + 1;
    if (i == j) {
      ans[i] = -1;
      continue;
    }
    ans[i] = ans[i + 1] + 2;
    while (~ans[i]) {
      if (get(i, i + ans[i] - 1) == get(j - ans[i] + 1, j)) break;
      ans[i] -= 2;
    }
  }
  for (int i = 1; i <= (n + 1) / 2; i++) printf("%d ", ans[i]);
  return 0;
}

```