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;
const int N = 2222;
int pos[N];
int f[N][N];
char s[N], p[N];
int sLen, pLen;
int main() {
  scanf("%s %s", s + 1, p + 1);
  sLen = strlen(s + 1);
  pLen = strlen(p + 1);
  for (int i = 1; i <= sLen; ++i) {
    if (s[i] == p[pLen]) {
      bool flag = true;
      int p1, p2;
      p1 = i, p2 = pLen;
      while (p2) {
        while (p1 && s[p1] != p[p2]) --p1;
        if (!p1) {
          flag = false;
          break;
        }
        --p2, --p1;
      }
      if (flag)
        pos[i] = p1;
      else
        pos[i] = -1;
    } else
      pos[i] = -1;
  }
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= sLen; ++i) {
    for (int j = 0; j <= i; ++j) {
      f[i][j] = f[i - 1][j];
      if (j) f[i][j] = max(f[i][j], f[i - 1][j - 1]);
      if (s[i] == p[pLen]) {
        int t = pos[i], d = i - t - pLen;
        if (t >= 0 && j - d >= 0 && j - d <= t) {
          f[i][j] = max(f[i][j], f[t][j - d] + 1);
        }
      }
    }
  }
  for (int i = 0; i <= sLen; ++i)
    printf("%d%c", f[sLen][i], i == sLen ? '\n' : ' ');
  return 0;
}
