Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a result. Then he calculates <image> that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s'. He wants to make this number as big as possible.

More formally, let's define <image> as maximum value of <image> over all s' that can be obtained by removing exactly x characters from s. Dreamoon wants to know <image> for all x from 0 to |s| where |s| denotes the length of string s.

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 2 000).

The second line of the input contains the string p (1 ≤ |p| ≤ 500).

Both strings will only consist of lower case English letters.

Output

Print |s| + 1 space-separated integers in a single line representing the <image> for all x from 0 to |s|.

Examples

Input

aaaaa
aa


Output

2 2 1 1 0 0


Input

axbaxxb
ab


Output

0 1 1 2 1 1 0 0

Note

For the first sample, the corresponding optimal values of s' after removal 0 through |s| = 5 characters from s are {"aaaaa", "aaaa", "aaa", "aa", "a", ""}. 

For the second sample, possible corresponding optimal values of s' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab", "a", ""}.
#include <bits/stdc++.h>
using namespace std;
long long inf = pow(10, 18);
long long modulo = pow(10, 9) + 7;
double eps = 1e-6;
const long long int mod = 1000000007;
int a, n;
string str, ptr;
int arr[2001];
long long dp[2001][2001];
long long finddp(int i, int j) {
  if (i >= (int)(str.size())) {
    if (j < 1)
      return 0;
    else
      return inf;
  }
  if (j < 1) return 0;
  if (dp[i][j] != -inf) return dp[i][j];
  if (str[i] == ptr[0] && arr[i] != -1) {
    dp[i][j] = arr[i] - i + 1 - (int)(ptr.size()) + finddp(arr[i] + 1, j - 1);
    return dp[i][j] = min(dp[i][j], finddp(i + 1, j));
  }
  return dp[i][j] = finddp(i + 1, j);
}
long long int min(long long int a, long long int b) { return a ? a < b : b; }
int main() {
  long long int a, f, l;
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> str >> ptr;
  for (int i = 0; i < (int)(str.size()); i++) {
    if (str[i] != ptr[0]) {
      arr[i] = -1;
      continue;
    }
    int counter = 1;
    for (int w = i + 1; w < (int)(str.size()); w++) {
      if (str[w] == ptr[counter]) {
        counter++;
      }
      if (counter >= (int)(ptr.size())) {
        arr[i] = w;
        break;
      }
    }
    if (counter < (int)(ptr.size())) {
      arr[i] = -1;
    }
  }
  for (int i = 0; i < 2001; i++)
    for (int j = 0; j < 2001; j++) dp[i][j] = -inf;
  long long arrnew[(int)(str.size()) + 1];
  for (int i = 0; i <= (int)(str.size()); i++) {
    arrnew[i] = (finddp(0, i));
  }
  for (int i = 0; i <= (int)(str.size()); i++) {
    int maxi = 4;
    for (int j = (int)(str.size()); j >= 0; j--) {
      long long temp = (int)(str.size()) - j * (int)(ptr.size());
      if (temp >= i && arrnew[j] <= (long long)i) {
        maxi = j;
        break;
      }
    }
    cout << maxi << endl;
  }
  cout << endl;
}
