#include <bits/stdc++.h>
using namespace std;
const int N = 1000006;
char s[N];
int n;
long long hs[N];
long long w[N], inv[N];
long long pot(long long b, long long e, long long m) {
  long long ans = 1;
  b %= m;
  while (e) {
    if (e & 1) ans = (ans * b) % m;
    b = (b * b) % m;
    e >>= 1;
  }
  return ans;
}
long long query(int i, int j) {
  return ((hs[j] + (i ? 1000000009 - hs[i - 1] : 0)) * inv[i]) % 1000000009;
}
bool ok(int i, int j, int sz) {
  return query(i - sz, i + sz) == query(j - sz, j + sz);
}
int ans[N];
int main() {
  w[0] = 1, inv[0] = 1;
  w[1] = 1009, inv[1] = pot(1009, 1000000009 - 2, 1000000009);
  for (int i = 2; i < N; i++)
    w[i] = (w[i - 1] * w[1]) % 1000000009,
    inv[i] = (inv[i - 1] * inv[1]) % 1000000009;
  scanf("%d %s", &n, s);
  for (int i = 0; i < n; i++) {
    hs[i] = (s[i] * w[i] + (i ? hs[i - 1] : 0)) % 1000000009;
  }
  memset(ans, -1, sizeof ans);
  for (int i = 0; i < n / 2; i++) {
    int j = n - 1 - i;
    if (s[i] != s[j]) continue;
    int L = 0, R = i, tmp = 0;
    while (L <= R) {
      int mid = (L + R) / 2;
      if (ok(i, j, mid))
        L = mid + 1, tmp = mid;
      else
        R = mid - 1;
    }
    ans[i - tmp] = 2 * tmp + 1;
  }
  for (int i = 0; i < n / 2; i++) {
    ans[i + 1] = max(ans[i + 1], ans[i] - 2);
  }
  printf("%d", ans[0]);
  for (int i = 1; i < n / 2; i++) printf(" %d", ans[i]);
  if (n & 1) printf(" -1");
  printf("\n");
}

### Prompt

Your task is to create 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 N = 1000006;
char s[N];
int n;
long long hs[N];
long long w[N], inv[N];
long long pot(long long b, long long e, long long m) {
  long long ans = 1;
  b %= m;
  while (e) {
    if (e & 1) ans = (ans * b) % m;
    b = (b * b) % m;
    e >>= 1;
  }
  return ans;
}
long long query(int i, int j) {
  return ((hs[j] + (i ? 1000000009 - hs[i - 1] : 0)) * inv[i]) % 1000000009;
}
bool ok(int i, int j, int sz) {
  return query(i - sz, i + sz) == query(j - sz, j + sz);
}
int ans[N];
int main() {
  w[0] = 1, inv[0] = 1;
  w[1] = 1009, inv[1] = pot(1009, 1000000009 - 2, 1000000009);
  for (int i = 2; i < N; i++)
    w[i] = (w[i - 1] * w[1]) % 1000000009,
    inv[i] = (inv[i - 1] * inv[1]) % 1000000009;
  scanf("%d %s", &n, s);
  for (int i = 0; i < n; i++) {
    hs[i] = (s[i] * w[i] + (i ? hs[i - 1] : 0)) % 1000000009;
  }
  memset(ans, -1, sizeof ans);
  for (int i = 0; i < n / 2; i++) {
    int j = n - 1 - i;
    if (s[i] != s[j]) continue;
    int L = 0, R = i, tmp = 0;
    while (L <= R) {
      int mid = (L + R) / 2;
      if (ok(i, j, mid))
        L = mid + 1, tmp = mid;
      else
        R = mid - 1;
    }
    ans[i - tmp] = 2 * tmp + 1;
  }
  for (int i = 0; i < n / 2; i++) {
    ans[i + 1] = max(ans[i + 1], ans[i] - 2);
  }
  printf("%d", ans[0]);
  for (int i = 1; i < n / 2; i++) printf(" %d", ans[i]);
  if (n & 1) printf(" -1");
  printf("\n");
}

```