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 s1[2005], s2[505];
int dp[2005][2005], del[2005];
int main() {
  scanf("%s%s", s1, s2);
  int l1 = strlen(s1), l2 = strlen(s2);
  for (int i = 0; i < l1; i++) {
    int cur = 0, j;
    for (j = i; j < l1; j++) {
      if (s1[j] == s2[cur]) {
        cur++;
        if (cur == l2) break;
      }
    }
    if (cur == l2)
      del[i] = j;
    else
      del[i] = -1;
  }
  for (int i = 0; i < l1; i++)
    for (int j = 0; j < l1 && 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 (del[i] != -1)
        dp[del[i] + 1][j + del[i] - i + 1 - l2] =
            max(dp[del[i] + 1][j + del[i] - i + 1 - l2], dp[i][j] + 1);
    }
  dp[l1][l1] = 0;
  for (int i = 0; i <= l1; i++) {
    printf("%d", dp[l1][i]);
    if (i != l1) printf(" ");
  }
  puts("");
  return 0;
}
