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;
int main() {
  string s;
  string p;
  cin >> s >> p;
  int n = s.size();
  int m = p.size();
  int next[n];
  int end[n];
  for (int i = 0; i < n; i++) {
    int counter = 0;
    int j = i;
    while (1) {
      if (j >= n) break;
      if (s[j] == p[0]) {
        counter++;
        break;
      }
      j++;
    }
    next[i] = j;
    j++;
    while (1) {
      if (counter >= m) {
        end[i] = j - 1;
        break;
      }
      if (j >= n) {
        next[i] = end[i] = n;
        break;
      }
      if (s[j] == p[counter]) {
        counter++;
      }
      j++;
    }
  }
  int dp[n + 1][n + 1];
  int mini[n + 1][n + 1];
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      mini[i][j] = dp[i][j] = 10000;
    }
  }
  for (int i = 0; i <= n; i++) {
    dp[n][0] = 0;
    mini[n][0] = 0;
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j <= n; j++) {
      if (next[i] < n) {
        dp[i][j] =
            min(dp[i][j], mini[end[i] + 1][j - 1] + end[i] - next[i] + 1 - m);
      }
      mini[i][j] = min(dp[i][j], mini[i + 1][j]);
    }
  }
  for (int i = 0; i <= n; i++) {
    int first = 0;
    int last = n + 1;
    int works = 0;
    while (first + 1 < last) {
      int j = (first + last) / 2;
      if (i < mini[0][j] || i > (n - m * j)) {
        last = j;
      } else {
        first = j;
      }
    }
    cout << first << " ";
  }
  cout << endl;
}
