#include <bits/stdc++.h>
using namespace std;
const int Maxn = 1000010;
const int inf = 2147483647;
const int base1 = 233, mod1 = 998244353;
const int base2 = 131, mod2 = 1000000007;
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * f;
}
int n, h1[Maxn], pow1[Maxn], h2[Maxn], pow2[Maxn], ans[Maxn];
char s[Maxn];
int g1(int l, int r) {
  return (h1[l] - (long long)h1[r + 1] * pow1[r - l + 1] % mod1 + mod1) % mod1;
}
int g2(int l, int r) {
  return (h2[l] - (long long)h2[r + 1] * pow2[r - l + 1] % mod2 + mod2) % mod2;
}
int main() {
  n = read();
  scanf("%s", s + 1);
  pow1[0] = 1;
  for (int i = 1; i <= n; i++) pow1[i] = (long long)pow1[i - 1] * base1 % mod1;
  pow2[0] = 1;
  for (int i = 1; i <= n; i++) pow2[i] = (long long)pow2[i - 1] * base2 % mod2;
  h1[n + 1] = 0;
  for (int i = n; i; i--)
    h1[i] = ((long long)h1[i + 1] * base1 % mod1 + s[i] - 'a') % mod1;
  h2[n + 1] = 0;
  for (int i = n; i; i--)
    h2[i] = ((long long)h2[i + 1] * base2 % mod2 + s[i] - 'a') % mod2;
  for (int i = 1; i <= (n >> 1); i++) {
    int l = 1, r = i;
    while (l <= r) {
      int mid = l + r >> 1;
      if (g1(i - mid + 1, i + mid - 1) ==
              g1(n - i + 1 - mid + 1, n - i + 1 + mid - 1) &&
          g2(i - mid + 1, i + mid - 1) ==
              g2(n - i + 1 - mid + 1, n - i + 1 + mid - 1))
        l = mid + 1;
      else
        r = mid - 1;
    }
    l--;
    ans[i - l + 1] = (l << 1) - 1;
  }
  for (int i = 1; i <= ((n + 1) >> 1); i++)
    ans[i] = max(ans[i], ans[i - 1] - 2),
    printf("%d ", (ans[i] == 0 ? -1 : ans[i]));
}

### 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 Maxn = 1000010;
const int inf = 2147483647;
const int base1 = 233, mod1 = 998244353;
const int base2 = 131, mod2 = 1000000007;
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * f;
}
int n, h1[Maxn], pow1[Maxn], h2[Maxn], pow2[Maxn], ans[Maxn];
char s[Maxn];
int g1(int l, int r) {
  return (h1[l] - (long long)h1[r + 1] * pow1[r - l + 1] % mod1 + mod1) % mod1;
}
int g2(int l, int r) {
  return (h2[l] - (long long)h2[r + 1] * pow2[r - l + 1] % mod2 + mod2) % mod2;
}
int main() {
  n = read();
  scanf("%s", s + 1);
  pow1[0] = 1;
  for (int i = 1; i <= n; i++) pow1[i] = (long long)pow1[i - 1] * base1 % mod1;
  pow2[0] = 1;
  for (int i = 1; i <= n; i++) pow2[i] = (long long)pow2[i - 1] * base2 % mod2;
  h1[n + 1] = 0;
  for (int i = n; i; i--)
    h1[i] = ((long long)h1[i + 1] * base1 % mod1 + s[i] - 'a') % mod1;
  h2[n + 1] = 0;
  for (int i = n; i; i--)
    h2[i] = ((long long)h2[i + 1] * base2 % mod2 + s[i] - 'a') % mod2;
  for (int i = 1; i <= (n >> 1); i++) {
    int l = 1, r = i;
    while (l <= r) {
      int mid = l + r >> 1;
      if (g1(i - mid + 1, i + mid - 1) ==
              g1(n - i + 1 - mid + 1, n - i + 1 + mid - 1) &&
          g2(i - mid + 1, i + mid - 1) ==
              g2(n - i + 1 - mid + 1, n - i + 1 + mid - 1))
        l = mid + 1;
      else
        r = mid - 1;
    }
    l--;
    ans[i - l + 1] = (l << 1) - 1;
  }
  for (int i = 1; i <= ((n + 1) >> 1); i++)
    ans[i] = max(ans[i], ans[i - 1] - 2),
    printf("%d ", (ans[i] == 0 ? -1 : ans[i]));
}

```