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 str, pattern;
int firstMatchEnd[2010];
int dp[2010][2010];
int INF = -100000000;
int solve(int nowIndex, int canBeDeleted) {
  int first = INF, second = INF, third = INF;
  if (nowIndex == (int)str.size() && canBeDeleted == 0) {
    return 0;
  } else if (nowIndex == (int)str.size()) {
    return INF;
  }
  if (dp[nowIndex][canBeDeleted] != -1) {
    return dp[nowIndex][canBeDeleted];
  }
  if (firstMatchEnd[nowIndex] != -1) {
    int nowL = firstMatchEnd[nowIndex] - nowIndex + 1;
    int changed = nowL - (int)pattern.size();
    if (canBeDeleted >= changed) {
      first = 1 + solve(firstMatchEnd[nowIndex] + 1, canBeDeleted - changed);
    }
  }
  second = solve(nowIndex + 1, canBeDeleted);
  third = solve(nowIndex + 1, canBeDeleted - 1);
  return dp[nowIndex][canBeDeleted] = max(first, max(second, third));
}
int main() {
  cin >> str >> pattern;
  memset(firstMatchEnd, -1, sizeof(firstMatchEnd));
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < (int)str.size(); i++) {
    int matched = 0;
    for (int j = i; j < (int)str.size(); j++) {
      if (str[j] == pattern[matched]) {
        matched++;
      }
      if (matched == (int)pattern.size()) {
        firstMatchEnd[i] = j;
        break;
      }
    }
  }
  cout << solve(0, 0);
  for (int i = 1; i <= (int)str.size(); i++) {
    cout << " " << solve(0, i);
  }
  cout << endl;
  return 0;
}
