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 s[2005], p[505];
int f[2005][2005], pos[2005];
int main() {
  memset(pos, -1, sizeof(pos));
  gets(s);
  gets(p);
  int str1 = strlen(s), str2 = strlen(p);
  for (int i = 0; i < str1; i++)
    if (s[i] == p[0]) {
      int cnt = 0;
      for (int j = i; j < str1; j++) {
        if (s[j] == p[cnt]) cnt++;
        if (cnt == str2) {
          pos[i] = j;
          break;
        }
      }
    }
  for (int i = 0; i <= str1; i++)
    for (int j = 1; j <= str1; j++) f[i][j] = 1 << 30;
  for (int i = str1 - 1; i >= 0; i--)
    if (pos[i] != -1) {
      for (int j = 1; j <= str1; j++)
        f[i][j] =
            min(f[i + 1][j], f[pos[i] + 1][j - 1] + (pos[i] - i + 1 - str2));
    } else {
      for (int j = 1; j <= str1; j++) f[i][j] = f[i + 1][j];
    }
  int cnt = 0;
  for (int i = 0; i <= str1; i++) {
    while (f[0][cnt] <= i) cnt++;
    while ((cnt - 1) * str2 > (str1 - i)) cnt--;
    printf("%d ", cnt - 1);
  }
  printf("\n");
  return 0;
}
