#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int MN = 1000010;
int po[MN];
int N;
char S[MN];
int psum[MN], mx[MN];
int calc(int l, int r) {
  return (psum[r] + mod - (l ? 1LL * psum[l - 1] * po[r - l + 1] % mod : 0)) %
         mod;
}
int main() {
  po[0] = 1;
  for (int i = 1; i < MN; i++) {
    po[i] = 1LL * po[i - 1] * 128 % mod;
  }
  scanf("%d %s", &N, S);
  for (int i = 0; i < N; i++) {
    psum[i] = S[i];
    if (i) {
      psum[i] += 1LL * psum[i - 1] * 128 % mod;
      psum[i] %= mod;
    }
  }
  memset(mx, -1, sizeof(mx));
  for (int i = 0; i < N / 2; i++) {
    int s = 0, e = i, p = -1;
    while (s <= e) {
      int m = (s + e) >> 1;
      if (calc(i - m, i + m) == calc(N - 1 - i - m, N - 1 - i + m)) {
        p = m;
        s = m + 1;
      } else
        e = m - 1;
    }
    if (p != -1) mx[i - p] = i;
  }
  int a = -1;
  for (int i = 0; i < N / 2; i++) {
    a = max(a, mx[i]);
    if (a < i)
      printf("-1 ");
    else
      printf("%d ", 2 * (a - i) + 1);
  }
  if (N % 2) printf("-1");
}

### Prompt

Your challenge is to write 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 int mod = 1e9 + 7;
const int MN = 1000010;
int po[MN];
int N;
char S[MN];
int psum[MN], mx[MN];
int calc(int l, int r) {
  return (psum[r] + mod - (l ? 1LL * psum[l - 1] * po[r - l + 1] % mod : 0)) %
         mod;
}
int main() {
  po[0] = 1;
  for (int i = 1; i < MN; i++) {
    po[i] = 1LL * po[i - 1] * 128 % mod;
  }
  scanf("%d %s", &N, S);
  for (int i = 0; i < N; i++) {
    psum[i] = S[i];
    if (i) {
      psum[i] += 1LL * psum[i - 1] * 128 % mod;
      psum[i] %= mod;
    }
  }
  memset(mx, -1, sizeof(mx));
  for (int i = 0; i < N / 2; i++) {
    int s = 0, e = i, p = -1;
    while (s <= e) {
      int m = (s + e) >> 1;
      if (calc(i - m, i + m) == calc(N - 1 - i - m, N - 1 - i + m)) {
        p = m;
        s = m + 1;
      } else
        e = m - 1;
    }
    if (p != -1) mx[i - p] = i;
  }
  int a = -1;
  for (int i = 0; i < N / 2; i++) {
    a = max(a, mx[i]);
    if (a < i)
      printf("-1 ");
    else
      printf("%d ", 2 * (a - i) + 1);
  }
  if (N % 2) printf("-1");
}

```