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 INF = 1000000001;
const double EPS = 10e-9;
int count_shortest(int i, string& str, string& wzo) {
  int j = i;
  int k = 0;
  while (k < ((int)(wzo).size()) && j < ((int)(str).size())) {
    if (str[j] == wzo[k]) {
      ++k;
      if (k == ((int)(wzo).size())) return j + 1;
    }
    ++j;
  }
  return j + 1;
}
int dp[2345][2345];
int main() {
  string str, wzo;
  int n, p, shortest[2345];
  cin >> str >> wzo;
  n = ((int)(str).size());
  p = ((int)(wzo).size());
  for (int i = 0; i < (n); ++i) shortest[i] = count_shortest(i, str, wzo);
  for (int i = 0; i < (n + 1); ++i) dp[n][i] = -INF;
  dp[n][0] = 0;
  for (int i = n - 1; i >= (0); --i) {
    for (int j = 0; j < (n + 1); ++j) {
      if (j > 0)
        dp[i][j] = max(dp[i + 1][j], dp[i + 1][j - 1]);
      else
        dp[i][j] = dp[i + 1][j];
      if (str[i] == wzo[0] && n >= shortest[i] && (j >= shortest[i] - i - p))
        dp[i][j] = max(dp[i][j], 1 + dp[shortest[i]][j - shortest[i] + i + p]);
    }
  }
  for (int j = 0; j < (n + 1); ++j) cout << dp[0][j] << " ";
  return 0;
}
