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;
template <typename C>
void ma(C& a, C b) {
  if (a < b) a = b;
}
template <typename C>
void mi(C& a, C b) {
  if (a > b) a = b;
}
int dp[2011][2011];
char s[2011], p[2011];
int n, m, q[2011][2011];
int main() {
  scanf("%s%s", s, p);
  n = 0;
  while (s[n]) n++;
  m = 0;
  while (p[m]) m++;
  for (int i = 0; i < n + 1; i++) q[0][i] = i;
  for (int i = 0; i < m; i++) {
    q[i + 1][0] = -1;
    for (int j = 0; j < n; j++) {
      if (p[i] == s[j])
        q[i + 1][j + 1] = q[i][j];
      else
        q[i + 1][j + 1] = q[i + 1][j];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < i + 2; j++) {
      dp[i + 1][j] = dp[i][j];
      if (j) ma(dp[i + 1][j], dp[i][j - 1]);
      if (q[m][i + 1] != -1) {
        int us = i - q[m][i + 1] + 1 - m;
        if (j >= us && q[m][i + 1] >= j - us)
          ma(dp[i + 1][j], dp[q[m][i + 1]][j - us] + 1);
      }
    }
  }
  for (int i = 0; i < n + 1; i++) printf("%d ", dp[n][i]);
  puts("");
}
