#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
const unsigned long long seed = 3121;
const long long seed1 = 233;
const int mo = 19260817;
int n, ans[maxn];
char s[maxn];
unsigned long long sum[maxn], p[maxn];
long long summ[maxn], p1[maxn];
void solve(int st, int ed) {
  int up = ans[st + 1] + 2;
  if (!(up & 1)) up--;
  if (st == ed) {
    return;
  }
  for (int mid = up; mid >= 1; mid -= 2) {
    unsigned long long sum1, sum2;
    sum1 = (sum[st + mid - 1] - sum[st - 1]) * p[ed - mid + 1 - st];
    sum2 = sum[ed] - sum[ed - mid];
    long long sum3, sum4;
    sum3 = (summ[st + mid - 1] - summ[st - 1] + mo) % mo *
           p1[ed - mid + 1 - st] % mo;
    sum4 = (summ[ed] - summ[ed - mid] + mo) % mo;
    if (sum1 == sum2 && sum3 == sum4) {
      ans[st] = mid;
      break;
    }
  }
}
int main() {
  scanf("%d%s", &n, s + 1);
  p[0] = 1;
  p1[0] = 1;
  memset(ans, -1, sizeof ans);
  for (int i = 1; i <= n; ++i) {
    p[i] = p[i - 1] * seed;
    p1[i] = p1[i - 1] * seed % mo;
    sum[i] = sum[i - 1] + p[i] * (unsigned long long)(s[i] - 'a');
    summ[i] = (summ[i - 1] + p1[i] * (s[i] - 'a')) % mo;
  }
  for (int i = n; i >= 1; --i)
    if (i <= n - i + 1) solve(i, n - i + 1);
  for (int i = 1; i <= n; ++i)
    if (i <= n - i + 1) printf("%d%c", ans[i], (i == n ? '\n' : ' '));
  return 0;
}

### 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 maxn = 1e6 + 10;
const unsigned long long seed = 3121;
const long long seed1 = 233;
const int mo = 19260817;
int n, ans[maxn];
char s[maxn];
unsigned long long sum[maxn], p[maxn];
long long summ[maxn], p1[maxn];
void solve(int st, int ed) {
  int up = ans[st + 1] + 2;
  if (!(up & 1)) up--;
  if (st == ed) {
    return;
  }
  for (int mid = up; mid >= 1; mid -= 2) {
    unsigned long long sum1, sum2;
    sum1 = (sum[st + mid - 1] - sum[st - 1]) * p[ed - mid + 1 - st];
    sum2 = sum[ed] - sum[ed - mid];
    long long sum3, sum4;
    sum3 = (summ[st + mid - 1] - summ[st - 1] + mo) % mo *
           p1[ed - mid + 1 - st] % mo;
    sum4 = (summ[ed] - summ[ed - mid] + mo) % mo;
    if (sum1 == sum2 && sum3 == sum4) {
      ans[st] = mid;
      break;
    }
  }
}
int main() {
  scanf("%d%s", &n, s + 1);
  p[0] = 1;
  p1[0] = 1;
  memset(ans, -1, sizeof ans);
  for (int i = 1; i <= n; ++i) {
    p[i] = p[i - 1] * seed;
    p1[i] = p1[i - 1] * seed % mo;
    sum[i] = sum[i - 1] + p[i] * (unsigned long long)(s[i] - 'a');
    summ[i] = (summ[i - 1] + p1[i] * (s[i] - 'a')) % mo;
  }
  for (int i = n; i >= 1; --i)
    if (i <= n - i + 1) solve(i, n - i + 1);
  for (int i = 1; i <= n; ++i)
    if (i <= n - i + 1) printf("%d%c", ans[i], (i == n ? '\n' : ' '));
  return 0;
}

```