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;
string S, P;
int getmatch(int pos) {
  int curr = 0;
  int st = pos;
  while (curr < P.size() && st < S.size()) {
    if (S[st] == P[curr]) {
      curr++;
      st++;
    } else {
      st++;
    }
  }
  if (curr == P.size())
    return st - 1;
  else
    return -1;
}
vector<int> dp[2222];
vector<pair<int, int> > intervals_ending_in[2222];
int main() {
  cin >> S >> P;
  int ct = 0;
  for (int j = 0; j < S.size(); j++) {
    int get_end = getmatch(j);
    if (get_end != -1) {
      intervals_ending_in[get_end + 1].push_back(
          make_pair(j + 1, get_end - j + 1 - (int)P.size()));
    }
  }
  for (int j = 0; j <= S.size(); j++) dp[j].resize(j + 1);
  dp[0][0] = 0;
  for (int j = 1; j <= S.size(); j++) {
    dp[j][j] = 0;
    for (int numtake = 0; numtake <= j - 1; numtake++) {
      dp[j][numtake] =
          max((numtake > 0 ? dp[j - 1][numtake - 1] : 0), dp[j - 1][numtake]);
      for (int k = 0; k < intervals_ending_in[j].size(); k++) {
        int st = intervals_ending_in[j][k].first;
        int sz = intervals_ending_in[j][k].second;
        if (numtake >= sz && dp[st - 1].size() > numtake - sz) {
          dp[j][numtake] = max(dp[j][numtake], dp[st - 1][numtake - sz] + 1);
        }
      }
    }
  }
  for (int j = 0; j <= S.size(); j++) {
    int maxval = 0;
    for (int en = 1; en <= S.size(); en++) {
      if (dp[en].size() >= j + 1) {
        maxval = max(maxval, dp[en][j]);
      }
    }
    printf("%d ", maxval);
  }
  printf("\n");
}
