#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const long long MOD = 1e9 + 7;
const long long Base = 19260817;
int n;
char s[N];
long long Hash[N], v[N];
int ans[N];
inline int check(int s, int t, int x) {
  if (s + x - 1 >= t || t - x + 1 <= s) return 0;
  long long t1 = (v[t] - v[t - x] * Hash[x]) % MOD;
  long long t2 = (v[s + x - 1] - v[s - 1] * Hash[x]) % MOD;
  if (t1 < 0) t1 += MOD;
  if (t2 < 0) t2 += MOD;
  return t1 == t2;
}
int main() {
  scanf("%d", &n);
  scanf(" %s", s + 1);
  Hash[0] = 1;
  for (int i = 1; i <= n; i++) {
    Hash[i] = Hash[i - 1] * Base % MOD;
    v[i] = (v[i - 1] * Base + s[i]) % MOD;
  }
  memset(ans, -1, sizeof(ans));
  for (int i = n + 1 >> 1; i; --i) {
    int l = i, r = n + 1 - i;
    if (l == r) {
      ans[i] = -1;
      continue;
    }
    for (int j = ans[i + 1] + 2; ~j; j -= 2) {
      if (check(l, r, j)) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= (n + 1) / 2; i++) printf("%d\t", ans[i]);
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
const int N = 1e6 + 5;
const long long MOD = 1e9 + 7;
const long long Base = 19260817;
int n;
char s[N];
long long Hash[N], v[N];
int ans[N];
inline int check(int s, int t, int x) {
  if (s + x - 1 >= t || t - x + 1 <= s) return 0;
  long long t1 = (v[t] - v[t - x] * Hash[x]) % MOD;
  long long t2 = (v[s + x - 1] - v[s - 1] * Hash[x]) % MOD;
  if (t1 < 0) t1 += MOD;
  if (t2 < 0) t2 += MOD;
  return t1 == t2;
}
int main() {
  scanf("%d", &n);
  scanf(" %s", s + 1);
  Hash[0] = 1;
  for (int i = 1; i <= n; i++) {
    Hash[i] = Hash[i - 1] * Base % MOD;
    v[i] = (v[i - 1] * Base + s[i]) % MOD;
  }
  memset(ans, -1, sizeof(ans));
  for (int i = n + 1 >> 1; i; --i) {
    int l = i, r = n + 1 - i;
    if (l == r) {
      ans[i] = -1;
      continue;
    }
    for (int j = ans[i + 1] + 2; ~j; j -= 2) {
      if (check(l, r, j)) {
        ans[i] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= (n + 1) / 2; i++) printf("%d\t", ans[i]);
  return 0;
}

```