#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int mod = 1000000123;
const long long p = 29;
const int INF = 1000000000;
int n, HALF;
long long power[1000005], h[1000005], tmpHash;
string str;
int result[1000005], res[1000005];
set<pair<int, int> > s;
int getHash(int l, int r);
int main() {
  int leftPos, rightPos, l, r, mid;
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> str;
  HALF = (n + 1) >> 1;
  str = "$" + str;
  power[0] = 1;
  for (int i = 1; i <= n; i++) power[i] = power[i - 1] * p % mod;
  for (int i = 1; i <= n; i++) h[i] = (h[i - 1] * p + (str[i] - 'a' + 1)) % mod;
  fill(result, result + n, -1);
  for (int i = 1; i <= HALF; i++) {
    leftPos = i;
    rightPos = n + 1 - i;
    l = 0;
    r = i - 1;
    if (str[leftPos] != str[rightPos]) continue;
    for (int iter = 0; (iter < 22) && (l <= r); iter++) {
      mid = (l + r) >> 1;
      if (getHash(leftPos - mid, leftPos + mid) ==
          getHash(rightPos - mid, rightPos + mid)) {
        result[i] = max(result[i], mid);
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
  }
  for (int i = HALF; i; i--) {
    s.insert({i, result[i]});
    s.erase({n + 1 - i, result[n + 1 - i]});
    while (!s.empty() && (s.rbegin()->first - i > s.rbegin()->second))
      s.erase(*s.rbegin());
    if (s.empty())
      res[i] = -1;
    else
      res[i] = (s.rbegin()->first - i) * 2 + 1;
  }
  for (int i = 1; i <= HALF; i++) cout << res[i] << " ";
  cout << "\n";
  return 0;
}
int getHash(int l, int r) {
  tmpHash = (h[r] - h[l - 1] * power[r - l + 1]) % mod;
  if (tmpHash < 0) tmpHash += mod;
  return tmpHash;
}

### Prompt

Please formulate 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int mod = 1000000123;
const long long p = 29;
const int INF = 1000000000;
int n, HALF;
long long power[1000005], h[1000005], tmpHash;
string str;
int result[1000005], res[1000005];
set<pair<int, int> > s;
int getHash(int l, int r);
int main() {
  int leftPos, rightPos, l, r, mid;
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> str;
  HALF = (n + 1) >> 1;
  str = "$" + str;
  power[0] = 1;
  for (int i = 1; i <= n; i++) power[i] = power[i - 1] * p % mod;
  for (int i = 1; i <= n; i++) h[i] = (h[i - 1] * p + (str[i] - 'a' + 1)) % mod;
  fill(result, result + n, -1);
  for (int i = 1; i <= HALF; i++) {
    leftPos = i;
    rightPos = n + 1 - i;
    l = 0;
    r = i - 1;
    if (str[leftPos] != str[rightPos]) continue;
    for (int iter = 0; (iter < 22) && (l <= r); iter++) {
      mid = (l + r) >> 1;
      if (getHash(leftPos - mid, leftPos + mid) ==
          getHash(rightPos - mid, rightPos + mid)) {
        result[i] = max(result[i], mid);
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
  }
  for (int i = HALF; i; i--) {
    s.insert({i, result[i]});
    s.erase({n + 1 - i, result[n + 1 - i]});
    while (!s.empty() && (s.rbegin()->first - i > s.rbegin()->second))
      s.erase(*s.rbegin());
    if (s.empty())
      res[i] = -1;
    else
      res[i] = (s.rbegin()->first - i) * 2 + 1;
  }
  for (int i = 1; i <= HALF; i++) cout << res[i] << " ";
  cout << "\n";
  return 0;
}
int getHash(int l, int r) {
  tmpHash = (h[r] - h[l - 1] * power[r - l + 1]) % mod;
  if (tmpHash < 0) tmpHash += mod;
  return tmpHash;
}

```