#include <bits/stdc++.h>
const long long INF = (long long)0x3f3f3f3f3f3f3f, MAX = 9e18, MIN = -9e18;
const double PI = acos(-1.0);
const double eps = 1e-6;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
const int maxm = 100 + 10;
using namespace std;
int n, m, k;
int a[maxn];
string s;
long long hashs[maxn];
long long mul[maxn];
bool judge(int l, int r, int L, int R) {
  int len = r - l + 1;
  long long res1 = (hashs[r] - hashs[l - 1] * mul[len]) % mod;
  if (res1 < 0) res1 += mod;
  long long res2 = (hashs[R] - hashs[L - 1] * mul[len]) % mod;
  if (res2 < 0) res2 += mod;
  return (res1 == res2);
}
int ans[maxn];
void run_case() {
  cin >> n;
  cin >> s;
  s = " " + s;
  mul[0] = 1;
  long long bas = 131;
  for (int i = 1; i <= n; i++) mul[i] = (mul[i - 1] * bas) % mod;
  for (int i = 1; i <= n; i++) {
    hashs[i] = (hashs[i - 1] * bas + (s[i] - 'a')) % mod;
  }
  for (int i = 1; i <= (n) / 2; i++) {
    int l = 1;
    int r = i;
    int len = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (judge(i - mid + 1, i + mid - 1, n - i + 1 - mid + 1,
                n - i + 1 + mid - 1)) {
        l = mid + 1;
        len = mid;
      } else {
        r = mid - 1;
      }
    }
    len = l - 1;
    ans[i - len + 1] = 2 * len - 1;
  }
  for (int i = 1; i <= (n + 1) / 2; i++) {
    ans[i] = max(ans[i], ans[i - 1] - 2);
    if (ans[i] == 0)
      cout << "-1 ";
    else
      cout << ans[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  run_case();
}

### Prompt

Please provide a cpp coded solution 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>
const long long INF = (long long)0x3f3f3f3f3f3f3f, MAX = 9e18, MIN = -9e18;
const double PI = acos(-1.0);
const double eps = 1e-6;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
const int maxm = 100 + 10;
using namespace std;
int n, m, k;
int a[maxn];
string s;
long long hashs[maxn];
long long mul[maxn];
bool judge(int l, int r, int L, int R) {
  int len = r - l + 1;
  long long res1 = (hashs[r] - hashs[l - 1] * mul[len]) % mod;
  if (res1 < 0) res1 += mod;
  long long res2 = (hashs[R] - hashs[L - 1] * mul[len]) % mod;
  if (res2 < 0) res2 += mod;
  return (res1 == res2);
}
int ans[maxn];
void run_case() {
  cin >> n;
  cin >> s;
  s = " " + s;
  mul[0] = 1;
  long long bas = 131;
  for (int i = 1; i <= n; i++) mul[i] = (mul[i - 1] * bas) % mod;
  for (int i = 1; i <= n; i++) {
    hashs[i] = (hashs[i - 1] * bas + (s[i] - 'a')) % mod;
  }
  for (int i = 1; i <= (n) / 2; i++) {
    int l = 1;
    int r = i;
    int len = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (judge(i - mid + 1, i + mid - 1, n - i + 1 - mid + 1,
                n - i + 1 + mid - 1)) {
        l = mid + 1;
        len = mid;
      } else {
        r = mid - 1;
      }
    }
    len = l - 1;
    ans[i - len + 1] = 2 * len - 1;
  }
  for (int i = 1; i <= (n + 1) / 2; i++) {
    ans[i] = max(ans[i], ans[i - 1] - 2);
    if (ans[i] == 0)
      cout << "-1 ";
    else
      cout << ans[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  run_case();
}

```