#include <bits/stdc++.h>
using namespace std;
int n;
long long f[2000005], sum[2000005], ans[1000002];
const int mod1 = 998244353;
char s[2000005];
int main() {
  scanf("%d%s", &n, s + 1);
  sum[0] = 1;
  for (int i = 1; i <= n; ++i) {
    f[i] = (f[i - 1] * 131 + s[i]) % mod1;
    sum[i] = (sum[i - 1] * 131) % mod1;
  }
  long long cnt = (n + 1) >> 1;
  long long l, r;
  if (n & 1) {
    ans[cnt] = -1;
    l = r = (n + 1) >> 1;
  } else {
    l = n >> 1;
    r = l + 1;
    ans[cnt] = s[l] == s[r] ? 1 : -1;
  }
  long long p, q;
  while (cnt--) {
    ans[cnt] = ans[cnt + 1] + 2;
    if (!cnt) break;
    l--;
    r++;
    p = -1;
    q = 1;
    while (ans[cnt] != -1) {
      p = (f[l + ans[cnt] - 1] - f[l - 1] * sum[ans[cnt]] % mod1 + mod1) % mod1;
      q = (f[r] - f[r - ans[cnt]] * sum[ans[cnt]] % mod1 + mod1) % mod1;
      if (p == q) break;
      ans[cnt] -= 2;
    }
  }
  for (int i = 1; i <= (n + 1) >> 1; ++i) printf("%lld ", ans[i]);
}

### 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;
int n;
long long f[2000005], sum[2000005], ans[1000002];
const int mod1 = 998244353;
char s[2000005];
int main() {
  scanf("%d%s", &n, s + 1);
  sum[0] = 1;
  for (int i = 1; i <= n; ++i) {
    f[i] = (f[i - 1] * 131 + s[i]) % mod1;
    sum[i] = (sum[i - 1] * 131) % mod1;
  }
  long long cnt = (n + 1) >> 1;
  long long l, r;
  if (n & 1) {
    ans[cnt] = -1;
    l = r = (n + 1) >> 1;
  } else {
    l = n >> 1;
    r = l + 1;
    ans[cnt] = s[l] == s[r] ? 1 : -1;
  }
  long long p, q;
  while (cnt--) {
    ans[cnt] = ans[cnt + 1] + 2;
    if (!cnt) break;
    l--;
    r++;
    p = -1;
    q = 1;
    while (ans[cnt] != -1) {
      p = (f[l + ans[cnt] - 1] - f[l - 1] * sum[ans[cnt]] % mod1 + mod1) % mod1;
      q = (f[r] - f[r - ans[cnt]] * sum[ans[cnt]] % mod1 + mod1) % mod1;
      if (p == q) break;
      ans[cnt] -= 2;
    }
  }
  for (int i = 1; i <= (n + 1) >> 1; ++i) printf("%lld ", ans[i]);
}

```