#include <bits/stdc++.h>
using namespace std;
int n, cnt = 1, p, fail[2000005], dfail[2000005], nxt[2000005][26],
       len[2000005], pos[2000005];
char s[1000005], a[2000005];
inline void extend(int x) {
  int c = a[x] - 'a';
  while (a[x - len[p] - 1] != a[x]) p = fail[p];
  if (!nxt[p][c]) {
    int cur = ++cnt, k = fail[p];
    len[cur] = len[p] + 2;
    while (a[x - len[k] - 1] != a[x]) k = fail[k];
    fail[cur] = nxt[k][c];
    if (len[fail[cur]] % 2 == 0 && (len[fail[cur]] / 2) % 2 == 1)
      dfail[cur] = fail[cur];
    else
      dfail[cur] = dfail[fail[cur]];
    nxt[p][c] = cur;
  }
  p = nxt[p][c];
}
int main(void) {
  register int i, j, t;
  scanf("%d", &n), scanf("%s", s + 1);
  for (i = 1; i <= n; ++i) a[i * 2 - 1] = s[i], a[i * 2] = s[n - i + 1];
  n <<= 1, fail[0] = fail[1] = 1, len[1] = -1;
  for (i = 1; i <= n; ++i) extend(i), pos[i] = p;
  memset(nxt, 0, sizeof(nxt));
  for (i = 1; i <= n; ++i) {
    nxt[i][0] = dfail[i];
    for (j = 1; j <= 20; ++j) nxt[i][j] = nxt[nxt[i][j - 1]][j - 1];
  }
  for (i = n, t = n; i > n / 2; i -= 2, t -= 4) {
    int o = pos[i];
    for (j = 20; ~j; --j)
      if (len[nxt[o][j]] >= t) o = nxt[o][j];
    o = dfail[o];
    printf("%d ", o ? len[o] / 2 : -1);
  }
  return 0;
}

### 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;
int n, cnt = 1, p, fail[2000005], dfail[2000005], nxt[2000005][26],
       len[2000005], pos[2000005];
char s[1000005], a[2000005];
inline void extend(int x) {
  int c = a[x] - 'a';
  while (a[x - len[p] - 1] != a[x]) p = fail[p];
  if (!nxt[p][c]) {
    int cur = ++cnt, k = fail[p];
    len[cur] = len[p] + 2;
    while (a[x - len[k] - 1] != a[x]) k = fail[k];
    fail[cur] = nxt[k][c];
    if (len[fail[cur]] % 2 == 0 && (len[fail[cur]] / 2) % 2 == 1)
      dfail[cur] = fail[cur];
    else
      dfail[cur] = dfail[fail[cur]];
    nxt[p][c] = cur;
  }
  p = nxt[p][c];
}
int main(void) {
  register int i, j, t;
  scanf("%d", &n), scanf("%s", s + 1);
  for (i = 1; i <= n; ++i) a[i * 2 - 1] = s[i], a[i * 2] = s[n - i + 1];
  n <<= 1, fail[0] = fail[1] = 1, len[1] = -1;
  for (i = 1; i <= n; ++i) extend(i), pos[i] = p;
  memset(nxt, 0, sizeof(nxt));
  for (i = 1; i <= n; ++i) {
    nxt[i][0] = dfail[i];
    for (j = 1; j <= 20; ++j) nxt[i][j] = nxt[nxt[i][j - 1]][j - 1];
  }
  for (i = n, t = n; i > n / 2; i -= 2, t -= 4) {
    int o = pos[i];
    for (j = 20; ~j; --j)
      if (len[nxt[o][j]] >= t) o = nxt[o][j];
    o = dfail[o];
    printf("%d ", o ? len[o] / 2 : -1);
  }
  return 0;
}

```