#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, MOD1 = 1e9 + 1337, MOD2 = 999888777;
int n, m, h1[N], h2[N], ans[N], p1[N], p2[N];
string s;
inline pair<int, int> get_hash(int l, int r) {
  int val1 = (h1[r] - h1[l - 1] + MOD1) % MOD1 * 1LL * p1[N - r] % MOD1,
      val2 = (h2[r] - h2[l - 1] + MOD2) % MOD2 * 1LL * p2[N - r] % MOD2;
  return make_pair(val1, val2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  p1[0] = p2[0] = 1;
  for (int i = 1; i < N; i++) {
    p1[i] = 1LL * p1[i - 1] * 59 % MOD1;
    p2[i] = 1LL * p2[i - 1] * 47 % MOD2;
  }
  cin >> n >> s;
  s = '#' + s;
  m = (n + 1) / 2;
  for (int i = 1; i <= n; i++) {
    h1[i] = (h1[i - 1] + 1LL * s[i] * p1[i]) % MOD1;
    h2[i] = (h2[i - 1] + 1LL * s[i] * p2[i]) % MOD2;
  }
  memset(ans, -1, sizeof(ans));
  for (int i = 1; i <= n / 2; i++) {
    int l = 1, r = n;
    while (r - l > 1) {
      int mid = (r + l) >> 1;
      int len = mid;
      int l1 = i - len + 1, r1 = i + len - 1, l2 = (n - i + 1) - len + 1,
          r2 = (n - i + 1) + len - 1;
      if (l1 < 0 || l2 < 0 || r1 > n || r2 > n) {
        r = mid;
        continue;
      }
      if (get_hash(l1, r1) == get_hash(l2, r2)) {
        l = mid;
      } else {
        r = mid;
      }
    }
    int len, l1, r1, l2, r2;
    len = r;
    l1 = i - len + 1;
    r1 = i + len - 1;
    l2 = (n - i + 1) - len + 1;
    r2 = (n - i + 1) + len - 1;
    if (!(l1 < 0 || l2 < 0 || r1 > n || r2 > n) &&
        get_hash(l1, r1) == get_hash(l2, r2)) {
      ans[i - len + 1] = max(ans[i - len + 1], 2 * len - 1);
    }
    len = l;
    l1 = i - len + 1;
    r1 = i + len - 1;
    l2 = (n - i + 1) - len + 1;
    r2 = (n - i + 1) + len - 1;
    if (!(l1 < 0 || l2 < 0 || r1 > n || r2 > n) &&
        get_hash(l1, r1) == get_hash(l2, r2)) {
      ans[i - len + 1] = max(ans[i - len + 1], 2 * len - 1);
    }
  }
  for (int i = 1; i <= m; i++) {
    ans[i] = max(ans[i], ans[i - 1] - 2);
    cout << ans[i] << " ";
  }
}

### Prompt

Please create a solution in Cpp 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 int N = 1e6 + 5, MOD1 = 1e9 + 1337, MOD2 = 999888777;
int n, m, h1[N], h2[N], ans[N], p1[N], p2[N];
string s;
inline pair<int, int> get_hash(int l, int r) {
  int val1 = (h1[r] - h1[l - 1] + MOD1) % MOD1 * 1LL * p1[N - r] % MOD1,
      val2 = (h2[r] - h2[l - 1] + MOD2) % MOD2 * 1LL * p2[N - r] % MOD2;
  return make_pair(val1, val2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  p1[0] = p2[0] = 1;
  for (int i = 1; i < N; i++) {
    p1[i] = 1LL * p1[i - 1] * 59 % MOD1;
    p2[i] = 1LL * p2[i - 1] * 47 % MOD2;
  }
  cin >> n >> s;
  s = '#' + s;
  m = (n + 1) / 2;
  for (int i = 1; i <= n; i++) {
    h1[i] = (h1[i - 1] + 1LL * s[i] * p1[i]) % MOD1;
    h2[i] = (h2[i - 1] + 1LL * s[i] * p2[i]) % MOD2;
  }
  memset(ans, -1, sizeof(ans));
  for (int i = 1; i <= n / 2; i++) {
    int l = 1, r = n;
    while (r - l > 1) {
      int mid = (r + l) >> 1;
      int len = mid;
      int l1 = i - len + 1, r1 = i + len - 1, l2 = (n - i + 1) - len + 1,
          r2 = (n - i + 1) + len - 1;
      if (l1 < 0 || l2 < 0 || r1 > n || r2 > n) {
        r = mid;
        continue;
      }
      if (get_hash(l1, r1) == get_hash(l2, r2)) {
        l = mid;
      } else {
        r = mid;
      }
    }
    int len, l1, r1, l2, r2;
    len = r;
    l1 = i - len + 1;
    r1 = i + len - 1;
    l2 = (n - i + 1) - len + 1;
    r2 = (n - i + 1) + len - 1;
    if (!(l1 < 0 || l2 < 0 || r1 > n || r2 > n) &&
        get_hash(l1, r1) == get_hash(l2, r2)) {
      ans[i - len + 1] = max(ans[i - len + 1], 2 * len - 1);
    }
    len = l;
    l1 = i - len + 1;
    r1 = i + len - 1;
    l2 = (n - i + 1) - len + 1;
    r2 = (n - i + 1) + len - 1;
    if (!(l1 < 0 || l2 < 0 || r1 > n || r2 > n) &&
        get_hash(l1, r1) == get_hash(l2, r2)) {
      ans[i - len + 1] = max(ans[i - len + 1], 2 * len - 1);
    }
  }
  for (int i = 1; i <= m; i++) {
    ans[i] = max(ans[i], ans[i - 1] - 2);
    cout << ans[i] << " ";
  }
}

```