#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000010;
int ans[maxn], n;
char s[maxn];
int a[maxn], b[maxn], c[maxn], d[maxn];
const int MOD1 = 993258975, MOD2 = 934384734, base1 = 233, base2 = 197;
bool check(int l, int r, int L, int R) {
  int x = r - l + 1;
  int ans1 = (a[r] - 1ll * a[l - 1] * c[x] % MOD1 + MOD1) % MOD1;
  int ans2 = (a[R] - 1ll * a[L - 1] * c[x] % MOD1 + MOD1) % MOD1;
  if (ans1 != ans2) return 0;
  ans1 = (b[r] - 1ll * b[l - 1] * d[x] % MOD2 + MOD2) % MOD2;
  ans2 = (b[R] - 1ll * b[L - 1] * d[x] % MOD2 + MOD2) % MOD2;
  if (ans1 != ans2) return 0;
  return 1;
}
int main() {
  scanf("%d%s", &n, s + 1);
  c[0] = d[0] = 1;
  for (int i = 1; i <= n; i++)
    a[i] = (1ll * a[i - 1] * base1 + s[i]) % MOD1,
    b[i] = (1ll * b[i - 1] * base2 + s[i]) % MOD2;
  for (int i = 1; i <= n; i++)
    c[i] = 1ll * c[i - 1] * base1 % MOD1, d[i] = 1ll * d[i - 1] * base2 % MOD2;
  memset(ans, -1, sizeof(ans));
  for (int i = 1; i <= n / 2; i++) {
    int l = 1, r = i + 1, mid;
    while (l < r) {
      mid = (l + r) >> 1;
      if (check(i - mid + 1, i + mid - 1, n - i + 1 - mid + 1,
                n - i + 1 + mid - 1))
        l = mid + 1;
      else
        r = mid;
    }
    l--;
    ans[i - l + 1] = max(ans[i - l + 1], 2 * l - 1);
  }
  for (int i = 1; i <= n / 2; i++) {
    ans[i] = max(ans[i], ans[i - 1] - 2);
    printf("%d ", ans[i]);
  }
  if (n & 1) printf("-1");
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = 1000010;
int ans[maxn], n;
char s[maxn];
int a[maxn], b[maxn], c[maxn], d[maxn];
const int MOD1 = 993258975, MOD2 = 934384734, base1 = 233, base2 = 197;
bool check(int l, int r, int L, int R) {
  int x = r - l + 1;
  int ans1 = (a[r] - 1ll * a[l - 1] * c[x] % MOD1 + MOD1) % MOD1;
  int ans2 = (a[R] - 1ll * a[L - 1] * c[x] % MOD1 + MOD1) % MOD1;
  if (ans1 != ans2) return 0;
  ans1 = (b[r] - 1ll * b[l - 1] * d[x] % MOD2 + MOD2) % MOD2;
  ans2 = (b[R] - 1ll * b[L - 1] * d[x] % MOD2 + MOD2) % MOD2;
  if (ans1 != ans2) return 0;
  return 1;
}
int main() {
  scanf("%d%s", &n, s + 1);
  c[0] = d[0] = 1;
  for (int i = 1; i <= n; i++)
    a[i] = (1ll * a[i - 1] * base1 + s[i]) % MOD1,
    b[i] = (1ll * b[i - 1] * base2 + s[i]) % MOD2;
  for (int i = 1; i <= n; i++)
    c[i] = 1ll * c[i - 1] * base1 % MOD1, d[i] = 1ll * d[i - 1] * base2 % MOD2;
  memset(ans, -1, sizeof(ans));
  for (int i = 1; i <= n / 2; i++) {
    int l = 1, r = i + 1, mid;
    while (l < r) {
      mid = (l + r) >> 1;
      if (check(i - mid + 1, i + mid - 1, n - i + 1 - mid + 1,
                n - i + 1 + mid - 1))
        l = mid + 1;
      else
        r = mid;
    }
    l--;
    ans[i - l + 1] = max(ans[i - l + 1], 2 * l - 1);
  }
  for (int i = 1; i <= n / 2; i++) {
    ans[i] = max(ans[i], ans[i - 1] - 2);
    printf("%d ", ans[i]);
  }
  if (n & 1) printf("-1");
  return 0;
}

```