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 MaxN = 2000 + 5;
char s1[MaxN], s2[MaxN];
int dp[MaxN][MaxN];
int len1, len2;
int solve(int x) {
  if (x < len2) return MaxN;
  int a = x, b = len2, tmp = 0;
  while (a && b) {
    if (s1[a] == s2[b])
      a--, b--;
    else
      tmp++, a--;
  }
  if (b == 0)
    return tmp;
  else
    return MaxN;
}
int main() {
  scanf("%s %s", s1 + 1, s2 + 1);
  len1 = strlen(s1 + 1);
  len2 = strlen(s2 + 1);
  for (int i = 0; i <= len1; i++)
    for (int j = 0; j <= len1; j++)
      if (j > i) dp[i][j] = -3000;
  for (int i = 1; i <= len1; i++) {
    int x = solve(i);
    for (int k = 0; k <= len1; k++) dp[i][k] = max(dp[i][k], dp[i - 1][k]);
    for (int k = 0; k <= len1; k++)
      if (x <= k) dp[i][k] = max(dp[i][k], dp[i - x - len2][k - x] + 1);
  }
  for (int i = 0; i <= len1; i++) printf("%d ", dp[len1][i]);
  printf("\n");
  return 0;
}
