#include <bits/stdc++.h>
using namespace std;
const long long base = 1202;
const long long somod = 1e9 + 69;
const long long maxn = 1e6 + 6;
long long f[maxn], h[maxn], b[maxn];
long long gethash(long long l, long long r) {
  return (h[r] - h[l - 1] * b[r - l + 1] + somod * somod) % somod;
}
long long n;
string s;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("test.inp", "r"))
    freopen("test.inp", "r", stdin), freopen("test.out", "w", stdout);
  cin >> n >> s;
  fill_n(f, maxn, -1);
  b[0] = 1;
  for (long long i = 1; i <= n; i++) {
    b[i] = b[i - 1] * base % somod;
    h[i] = (h[i - 1] * base + s[i - 1]) % somod;
  }
  long long cur = -1;
  for (long long i = n / 2; i >= 1; i--) {
    long long j = n - i + 1;
    long long len = j - i;
    cur += 2;
    cur = min(cur, len);
    if (cur % 2 == 0) cur--;
    for (; cur > 0; cur -= 2) {
      if (gethash(i, i + cur - 1) == gethash(j - cur + 1, j)) {
        f[i] = cur;
        break;
      }
    }
  }
  for (long long i = 1; i <= (n + 1) / 2; i++) cout << f[i] << ' ';
}

### Prompt

Your task is to create 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 long long base = 1202;
const long long somod = 1e9 + 69;
const long long maxn = 1e6 + 6;
long long f[maxn], h[maxn], b[maxn];
long long gethash(long long l, long long r) {
  return (h[r] - h[l - 1] * b[r - l + 1] + somod * somod) % somod;
}
long long n;
string s;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("test.inp", "r"))
    freopen("test.inp", "r", stdin), freopen("test.out", "w", stdout);
  cin >> n >> s;
  fill_n(f, maxn, -1);
  b[0] = 1;
  for (long long i = 1; i <= n; i++) {
    b[i] = b[i - 1] * base % somod;
    h[i] = (h[i - 1] * base + s[i - 1]) % somod;
  }
  long long cur = -1;
  for (long long i = n / 2; i >= 1; i--) {
    long long j = n - i + 1;
    long long len = j - i;
    cur += 2;
    cur = min(cur, len);
    if (cur % 2 == 0) cur--;
    for (; cur > 0; cur -= 2) {
      if (gethash(i, i + cur - 1) == gethash(j - cur + 1, j)) {
        f[i] = cur;
        break;
      }
    }
  }
  for (long long i = 1; i <= (n + 1) / 2; i++) cout << f[i] << ' ';
}

```