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 = 2200;
int dp[N][N];
char s[N], p[N];
int main() {
  scanf("%s%s", s + 1, p);
  int ls = strlen(s + 1), lp = strlen(p);
  for (int i = 0; i <= ls; i++)
    for (int j = i + 1; j <= ls; j++) dp[i][j] = -N;
  for (int i = 1; s[i]; i++) {
    int era = 0;
    for (int j = 0; j <= i; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
    }
    int k = i, cur = lp - 1;
    for (; ~cur && k; k--) {
      if (s[k] == p[cur])
        cur--;
      else
        era++;
    }
    if (~cur) continue;
    for (int j = era; j <= i; j++) dp[i][j] = max(dp[i][j], dp[k][j - era] + 1);
  }
  for (int i = 0; i <= ls; i++) printf("%d ", dp[ls][i]);
  return 0;
}
