#include <bits/stdc++.h>
using namespace std;
const int N(1000005), P(1002928181);
int B[N];
int n, h[N];
char s[N];
int stk[N], *tot = stk;
int hsh(int l, int r) {
  int s = (h[r] - (long long)h[l - 1] * B[r - l + 1]) % P;
  return s < 0 ? s + P : s;
}
int main() {
  B[0] = 1;
  for (int i = 1; i < N; ++i) B[i] = B[i - 1] * 66233LL % P;
  scanf("%d%s", &n, s + 1);
  for (int i = 1; i <= n; ++i) h[i] = (h[i - 1] * 66233LL + s[i]) % P;
  if (n & 1) *tot++ = -1;
  int p = -1;
  for (int i = ((n & 1) ? (3) : (2)); i <= n; i += 2) {
    p += 2;
    while (~p && hsh(((n - i) >> 1) + 1, ((n - i) >> 1) + p) !=
                     hsh(n - ((n - i) >> 1) - p + 1, n - ((n - i) >> 1)))
      p -= 2;
    *tot++ = p;
  }
  while (tot != stk) printf("%d ", *--tot);
  putchar(10);
  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;
const int N(1000005), P(1002928181);
int B[N];
int n, h[N];
char s[N];
int stk[N], *tot = stk;
int hsh(int l, int r) {
  int s = (h[r] - (long long)h[l - 1] * B[r - l + 1]) % P;
  return s < 0 ? s + P : s;
}
int main() {
  B[0] = 1;
  for (int i = 1; i < N; ++i) B[i] = B[i - 1] * 66233LL % P;
  scanf("%d%s", &n, s + 1);
  for (int i = 1; i <= n; ++i) h[i] = (h[i - 1] * 66233LL + s[i]) % P;
  if (n & 1) *tot++ = -1;
  int p = -1;
  for (int i = ((n & 1) ? (3) : (2)); i <= n; i += 2) {
    p += 2;
    while (~p && hsh(((n - i) >> 1) + 1, ((n - i) >> 1) + p) !=
                     hsh(n - ((n - i) >> 1) - p + 1, n - ((n - i) >> 1)))
      p -= 2;
    *tot++ = p;
  }
  while (tot != stk) printf("%d ", *--tot);
  putchar(10);
  return 0;
}

```