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;
char str[2010], pattern[2010];
int dp[2010][2010];
int main(void) {
  scanf("%s", str);
  scanf("%s", pattern);
  int len1 = (int)strlen(str);
  int len2 = (int)strlen(pattern);
  for (int i = 1; i <= len1; ++i) {
    int k = len2, j = i;
    while (k > 0 && j > 0) {
      if (pattern[k - 1] == str[j - 1]) k--;
      j--;
    }
    if (k == 0) {
      int del = i - j - len2;
      for (int f = del; f <= i - len2; ++f)
        dp[i][f] = max(dp[i][f], dp[j][f - del] + 1);
    }
    for (int f = 0; f < i; ++f) dp[i][f] = max(dp[i][f], dp[i - 1][f]);
  }
  for (int i = 0; i <= len1; ++i) printf("%d ", dp[len1][i]);
  return 0;
}
