#include <bits/stdc++.h>
using namespace std;
static mt19937 __gen(chrono::steady_clock::now().time_since_epoch().count());
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
constexpr int inf = 0x3f3f3f3f;
constexpr ll lnf = 0x3f3f3f3f3f3f3f3f;
static const int Maxn = 1000005;
static const ull BASE = 29;
static const ull modtable[] = {
    1019260817,
    998244353,
    1000000007,
    1000000009,
};
static const ull mod = modtable[uniform_int_distribution<int>()(__gen) %
                                (sizeof(modtable) / sizeof(ull))];
int n;
char str[Maxn];
int ans[Maxn];
ull pw[Maxn], hs[Maxn];
inline ull get_hash(int l, int r) {
  if (l > r) return 0ULL;
  return (hs[r] - hs[l - 1] * pw[r - l + 1] % mod + mod) % mod;
}
signed main() {
  scanf("%d", &n);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = pw[i - 1] * BASE % mod;
  scanf("%s", str + 1);
  for (int i = 1; i <= n; ++i)
    hs[i] = (hs[i - 1] * BASE + (str[i] - 'a' + 1)) % mod;
  if (n & 1)
    ans[(n + 1) / 2] = -1;
  else
    ans[(n + 1) / 2] = ((str[n / 2] == str[n / 2 + 1]) ? 1 : -1);
  for (int i = (n - 1) / 2; i >= 1; --i) {
    int l = i, r = n - l + 1;
    for (int j = ans[i + 1] + 2; j >= 1; j -= 2) {
      ull lhs = get_hash(l, l + j - 1);
      ull rhs = get_hash(r - j + 1, r);
      if (lhs == rhs) {
        ans[i] = j;
        break;
      }
    }
    if (!ans[i]) ans[i] = -1;
  }
  for (int i = 1; i <= (n + 1) / 2; ++i)
    printf("%d%c", ans[i], " \n"[i == (n + 1) / 2]);
  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;
static mt19937 __gen(chrono::steady_clock::now().time_since_epoch().count());
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
constexpr int inf = 0x3f3f3f3f;
constexpr ll lnf = 0x3f3f3f3f3f3f3f3f;
static const int Maxn = 1000005;
static const ull BASE = 29;
static const ull modtable[] = {
    1019260817,
    998244353,
    1000000007,
    1000000009,
};
static const ull mod = modtable[uniform_int_distribution<int>()(__gen) %
                                (sizeof(modtable) / sizeof(ull))];
int n;
char str[Maxn];
int ans[Maxn];
ull pw[Maxn], hs[Maxn];
inline ull get_hash(int l, int r) {
  if (l > r) return 0ULL;
  return (hs[r] - hs[l - 1] * pw[r - l + 1] % mod + mod) % mod;
}
signed main() {
  scanf("%d", &n);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = pw[i - 1] * BASE % mod;
  scanf("%s", str + 1);
  for (int i = 1; i <= n; ++i)
    hs[i] = (hs[i - 1] * BASE + (str[i] - 'a' + 1)) % mod;
  if (n & 1)
    ans[(n + 1) / 2] = -1;
  else
    ans[(n + 1) / 2] = ((str[n / 2] == str[n / 2 + 1]) ? 1 : -1);
  for (int i = (n - 1) / 2; i >= 1; --i) {
    int l = i, r = n - l + 1;
    for (int j = ans[i + 1] + 2; j >= 1; j -= 2) {
      ull lhs = get_hash(l, l + j - 1);
      ull rhs = get_hash(r - j + 1, r);
      if (lhs == rhs) {
        ans[i] = j;
        break;
      }
    }
    if (!ans[i]) ans[i] = -1;
  }
  for (int i = 1; i <= (n + 1) / 2; ++i)
    printf("%d%c", ans[i], " \n"[i == (n + 1) / 2]);
  return 0;
}

```