#include <bits/stdc++.h>
using namespace std;
using ll = long long;
inline ll read() {
  ll x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const ll N = 2e6 + 10, M = 131, mod = 998244353;
ll l, r, cnt, ans[N >> 1], f[N], sum[N], n, p, q;
char str[N];
int main() {
  n = read();
  scanf("%s", str + 1);
  sum[0] = 1;
  for (ll i = 1; i <= n; i++) {
    f[i] = (f[i - 1] * M + str[i]) % mod;
    sum[i] = sum[i - 1] * M % mod;
  }
  cnt = n + 1 >> 1;
  if (n & 1) {
    ans[cnt] = -1;
    l = r = cnt;
  } else {
    l = n >> 1;
    r = l + 1;
    if (str[l] == str[r])
      ans[cnt] = 1;
    else
      ans[cnt] = -1;
  }
  while (cnt--) {
    ans[cnt] = ans[cnt + 1] + 2;
    if (!cnt) break;
    l--;
    r++;
    p = -1;
    q = 1;
    while (ans[cnt] != -1) {
      p = (f[l + ans[cnt] - 1] - f[l - 1] * sum[ans[cnt]] % mod + mod) % mod;
      q = (f[r] - f[r - ans[cnt]] * sum[ans[cnt]] % mod + mod) % mod;
      if (p == q) break;
      ans[cnt] -= 2;
    }
  }
  for (ll i = 1; i <= (n + 1 >> 1); i++) printf("%lld ", ans[i]);
  return 0;
}

### Prompt

Develop a solution in cpp 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>
using namespace std;
using ll = long long;
inline ll read() {
  ll x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const ll N = 2e6 + 10, M = 131, mod = 998244353;
ll l, r, cnt, ans[N >> 1], f[N], sum[N], n, p, q;
char str[N];
int main() {
  n = read();
  scanf("%s", str + 1);
  sum[0] = 1;
  for (ll i = 1; i <= n; i++) {
    f[i] = (f[i - 1] * M + str[i]) % mod;
    sum[i] = sum[i - 1] * M % mod;
  }
  cnt = n + 1 >> 1;
  if (n & 1) {
    ans[cnt] = -1;
    l = r = cnt;
  } else {
    l = n >> 1;
    r = l + 1;
    if (str[l] == str[r])
      ans[cnt] = 1;
    else
      ans[cnt] = -1;
  }
  while (cnt--) {
    ans[cnt] = ans[cnt + 1] + 2;
    if (!cnt) break;
    l--;
    r++;
    p = -1;
    q = 1;
    while (ans[cnt] != -1) {
      p = (f[l + ans[cnt] - 1] - f[l - 1] * sum[ans[cnt]] % mod + mod) % mod;
      q = (f[r] - f[r - ans[cnt]] * sum[ans[cnt]] % mod + mod) % mod;
      if (p == q) break;
      ans[cnt] -= 2;
    }
  }
  for (ll i = 1; i <= (n + 1 >> 1); i++) printf("%lld ", ans[i]);
  return 0;
}

```