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 n, k;
string s, p;
int dp[2011][2011];
int box[2011];
int len1, len2;
int main() {
  ios::sync_with_stdio(0);
  cin >> s >> p;
  s = '!' + s;
  p = '!' + p;
  len1 = s.size() - 1;
  len2 = p.size() - 1;
  for (int i = 1; i <= len1; i++) {
    if (s[i] == p[len2]) {
      int now = len2 - 1;
      if (now == 0) {
        box[i] = i;
        continue;
      }
      for (int j = i - 1; j >= 1; j--) {
        if (s[j] == p[now]) {
          now--;
        }
        if (now == 0) {
          box[i] = j;
          break;
        }
      }
    }
  }
  for (int i = 1; i <= len1; i++) {
    for (int j = 0; j <= i; j++) {
      if (j != 0) {
        dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1]);
      } else {
        dp[i][j] = dp[i - 1][j];
      }
      if (s[i] == p[len2] && box[i] > 0) {
        int a = i - box[i] + 1;
        int b = a - len2;
        if (j >= b && box[i] - 1 >= j - b) {
          dp[i][j] = max(dp[i][j], dp[box[i] - 1][j - b] + 1);
        }
      }
    }
  }
  for (int i = 0; i <= len1; i++) {
    printf("%d%c", dp[len1][i], " \n"[i == len1]);
  }
  return 0;
}
