#include <bits/stdc++.h>
using namespace std;
int n, p[2000005], Max[2000005], po[2000005], ha[2000005];
char c[2000005];
int xx(int x, int y) { return x + y - (x + y < 998244353 ? 0 : 998244353); }
bool Check(int p, int len) {
  int t = n - p + 1;
  if (t < p) swap(t, p);
  if ((long long)xx(ha[p + len - 1], 998244353 - ha[p - len]) * po[t - p] %
          998244353 ==
      xx(ha[t + len - 1], 998244353 - ha[t - len]))
    return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  scanf("%s", c + 1);
  po[0] = 1;
  for (int i = 1; i <= n; ++i)
    po[i] = (long long)po[i - 1] * 19990213 % 998244353;
  for (int i = 1; i <= n; ++i)
    ha[i] = xx(ha[i - 1], (long long)(c[i] - 'a' + 1) * po[i] % 998244353);
  for (int i = 1; i <= n / 2; ++i) {
    if (c[i] != c[n - i + 1]) continue;
    int L = 1, R = min(i, n - i + 1);
    while (L < R) {
      int mid = (L + R + 1) >> 1;
      if (Check(i, mid))
        L = mid;
      else
        R = mid - 1;
    }
    Max[i - L + 1] = max(Max[i - L + 1], i);
  }
  int mmax = 0;
  for (int i = 1; i <= (n + 1) / 2; ++i) {
    mmax = max(mmax, Max[i]);
    if (mmax < i)
      printf("-1 ");
    else
      printf("%d ", 2 * (mmax - i) + 1);
  }
  cout << endl;
  return 0;
}

### Prompt

Generate 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;
int n, p[2000005], Max[2000005], po[2000005], ha[2000005];
char c[2000005];
int xx(int x, int y) { return x + y - (x + y < 998244353 ? 0 : 998244353); }
bool Check(int p, int len) {
  int t = n - p + 1;
  if (t < p) swap(t, p);
  if ((long long)xx(ha[p + len - 1], 998244353 - ha[p - len]) * po[t - p] %
          998244353 ==
      xx(ha[t + len - 1], 998244353 - ha[t - len]))
    return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  scanf("%s", c + 1);
  po[0] = 1;
  for (int i = 1; i <= n; ++i)
    po[i] = (long long)po[i - 1] * 19990213 % 998244353;
  for (int i = 1; i <= n; ++i)
    ha[i] = xx(ha[i - 1], (long long)(c[i] - 'a' + 1) * po[i] % 998244353);
  for (int i = 1; i <= n / 2; ++i) {
    if (c[i] != c[n - i + 1]) continue;
    int L = 1, R = min(i, n - i + 1);
    while (L < R) {
      int mid = (L + R + 1) >> 1;
      if (Check(i, mid))
        L = mid;
      else
        R = mid - 1;
    }
    Max[i - L + 1] = max(Max[i - L + 1], i);
  }
  int mmax = 0;
  for (int i = 1; i <= (n + 1) / 2; ++i) {
    mmax = max(mmax, Max[i]);
    if (mmax < i)
      printf("-1 ");
    else
      printf("%d ", 2 * (mmax - i) + 1);
  }
  cout << endl;
  return 0;
}

```