#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1, c = getchar();
  while (c < 48) c == '-' && (f = -1), c = getchar();
  while (c > 47) x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const int mod = 1e9 + 7;
const int MAXN = 1000005;
long long h[MAXN], pw[MAXN], ans[MAXN];
long long n, len;
char s[MAXN];
inline long long gethash(int l, int r) {
  return (h[r] - h[l - 1] * pw[r - l + 1] % mod + mod) % mod;
}
int main(int argc, char const *argv[]) {
  n = read();
  scanf("%s", s + 1);
  for (int i = pw[0] = 1; i <= n; ++i) {
    pw[i] = pw[i - 1] * 131 % mod;
    h[i] = (h[i - 1] * 131 + s[i] - 'a') % mod;
  }
  memset(ans, -1, sizeof ans);
  len = (n + 1) >> 1;
  for (int i = len, j; i; --i)
    if ((j = n - i + 1, 1) && i == j)
      ans[i] = -1;
    else
      for (ans[i] = ans[i + 1] + 2; ~ans[i]; ans[i] -= 2)
        if (gethash(i, i + ans[i] - 1) == gethash(j - ans[i] + 1, j)) break;
  for (int i = 1; i <= len; ++i) printf("%lld%c", ans[i], " \n"[i == n]);
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
inline int read() {
  int x = 0, f = 1, c = getchar();
  while (c < 48) c == '-' && (f = -1), c = getchar();
  while (c > 47) x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const int mod = 1e9 + 7;
const int MAXN = 1000005;
long long h[MAXN], pw[MAXN], ans[MAXN];
long long n, len;
char s[MAXN];
inline long long gethash(int l, int r) {
  return (h[r] - h[l - 1] * pw[r - l + 1] % mod + mod) % mod;
}
int main(int argc, char const *argv[]) {
  n = read();
  scanf("%s", s + 1);
  for (int i = pw[0] = 1; i <= n; ++i) {
    pw[i] = pw[i - 1] * 131 % mod;
    h[i] = (h[i - 1] * 131 + s[i] - 'a') % mod;
  }
  memset(ans, -1, sizeof ans);
  len = (n + 1) >> 1;
  for (int i = len, j; i; --i)
    if ((j = n - i + 1, 1) && i == j)
      ans[i] = -1;
    else
      for (ans[i] = ans[i + 1] + 2; ~ans[i]; ans[i] -= 2)
        if (gethash(i, i + ans[i] - 1) == gethash(j - ans[i] + 1, j)) break;
  for (int i = 1; i <= len; ++i) printf("%lld%c", ans[i], " \n"[i == n]);
  return 0;
}

```