#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
const long long base = 131;
const long long mod = 1e9 + 7;
long long Hash[maxn], b[maxn];
char ch[maxn];
int n, t, res[maxn / 2];
inline int read() {
  int t = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    t = (t << 3) + (t << 1) + (ch ^ 48);
    ch = getchar();
  }
  return t;
}
inline void write(int t) {
  if (t < 0) {
    putchar('-');
    t = -t;
  }
  if (t > 9) {
    write(t / 10);
  }
  putchar(t % 10 + '0');
}
inline void init() {
  b[0] = 1;
  for (int i = 1; i <= n; i++) {
    b[i] = (b[i - 1] * base) % mod;
  }
  for (int i = 1; i <= n; i++) {
    Hash[i] = (Hash[i - 1] * base + ch[i] - 'a' + 1) % mod;
  }
}
inline long long get_hash(int l, int r) {
  return ((Hash[r] - Hash[l - 1] * b[r - l + 1]) % mod + mod) % mod;
}
inline void solve(int x) {
  int y = t + t - x;
  if (!(n & 1)) {
    y++;
  }
  for (int i = res[x + 1] + 2; i >= 1; i -= 2) {
    if (get_hash(x, x + i - 1) == get_hash(y - i + 1, y)) {
      res[x] = i;
      return;
    }
  }
  res[x] = -1;
  return;
}
int main() {
  n = read();
  scanf("%s", ch + 1);
  init();
  t = n / 2;
  if (n & 1) {
    t++;
    res[t] = -1;
  } else {
    if (ch[t] == ch[t + 1]) {
      res[t] = 1;
    } else {
      res[t] = -1;
    }
  }
  for (int i = t - 1; i > 0; i--) {
    solve(i);
  }
  for (int i = 1; i <= t; i++) {
    write(res[i]);
    if (i == t) {
      putchar(10);
    } else {
      putchar(' ');
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 + 5;
const long long base = 131;
const long long mod = 1e9 + 7;
long long Hash[maxn], b[maxn];
char ch[maxn];
int n, t, res[maxn / 2];
inline int read() {
  int t = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    t = (t << 3) + (t << 1) + (ch ^ 48);
    ch = getchar();
  }
  return t;
}
inline void write(int t) {
  if (t < 0) {
    putchar('-');
    t = -t;
  }
  if (t > 9) {
    write(t / 10);
  }
  putchar(t % 10 + '0');
}
inline void init() {
  b[0] = 1;
  for (int i = 1; i <= n; i++) {
    b[i] = (b[i - 1] * base) % mod;
  }
  for (int i = 1; i <= n; i++) {
    Hash[i] = (Hash[i - 1] * base + ch[i] - 'a' + 1) % mod;
  }
}
inline long long get_hash(int l, int r) {
  return ((Hash[r] - Hash[l - 1] * b[r - l + 1]) % mod + mod) % mod;
}
inline void solve(int x) {
  int y = t + t - x;
  if (!(n & 1)) {
    y++;
  }
  for (int i = res[x + 1] + 2; i >= 1; i -= 2) {
    if (get_hash(x, x + i - 1) == get_hash(y - i + 1, y)) {
      res[x] = i;
      return;
    }
  }
  res[x] = -1;
  return;
}
int main() {
  n = read();
  scanf("%s", ch + 1);
  init();
  t = n / 2;
  if (n & 1) {
    t++;
    res[t] = -1;
  } else {
    if (ch[t] == ch[t + 1]) {
      res[t] = 1;
    } else {
      res[t] = -1;
    }
  }
  for (int i = t - 1; i > 0; i--) {
    solve(i);
  }
  for (int i = 1; i <= t; i++) {
    write(res[i]);
    if (i == t) {
      putchar(10);
    } else {
      putchar(' ');
    }
  }
  return 0;
}

```