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, e;
int dp[2005][2005];
int ct[2000];
int n;
int cost(int p) {
  int i, j, k;
  for (i = 0, j = p; i < e.size(); i++) {
    for (; (j < s.size()) && (e[i] != s[j]); j++)
      ;
    if (j == s.size()) return -1;
    j++;
  }
  return j;
}
int fun(int p, int r) {
  if (dp[p][r] > -2) {
    return dp[p][r];
  }
  if (p == n) {
    if (r == 0) return 0;
    return -1;
  }
  if (r > n - p) {
    dp[p][r] = -1;
    return -1;
  }
  if (r == n - p) {
    dp[p][r] = 0;
    return 0;
  }
  int i, j, k = 0;
  if (s[p] == e[0]) {
    j = ct[p];
    if (j == -1) {
      dp[p][r] = 0;
      return 0;
    }
    i = j - p - e.size();
    if (i <= r) {
      i = fun(j, r - i);
      if (i >= 0) k = i + 1;
    }
  }
  j = fun(p + 1, r);
  if (j > k) k = j;
  if (r) {
    j = fun(p + 1, r - 1);
    if (j > k) k = j;
  }
  dp[p][r] = k;
  return k;
}
int main() {
  int i, j, k, ans;
  cin >> s >> e;
  n = s.size();
  for (i = 0; i <= 2000; i++)
    for (j = 0; j <= 2000; j++) dp[i][j] = -2;
  for (i = 0; i < s.size(); i++) {
    if (s[i] == e[0]) ct[i] = cost(i);
  }
  for (i = 0; i <= s.size(); i++) cout << fun(0, i) << " ";
  cout << endl;
  return 0;
}
