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 dp[2001][2001];
char s[2001];
char p[2001];
int a[2001];
int main() {
  gets(s);
  gets(p);
  int lenp = strlen(p), lens = strlen(s);
  for (int i = 0; i < lens; i++) {
    int j, ps = i;
    for (j = 0; j < lenp; j++) {
      while (ps < lens && p[j] != s[ps]) ps++;
      if (ps == lens)
        break;
      else
        ps++;
    }
    if (j == lenp)
      a[i] = ps - i;
    else
      a[i] = -1;
  }
  for (int i = 0; i < lens; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (a[i] != -1) {
        dp[i + a[i]][j + a[i] - lenp] =
            max(dp[i + a[i]][j + a[i] - lenp], dp[i][j] + 1);
      }
    }
  }
  for (int i = 0; i <= lens; i++) printf("%d ", dp[lens][i]);
  return 0;
}
