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;
int dp[2001][2001], a[2001];
int l, l1;
string s, p;
int rec(int c, int x) {
  if (dp[c][x] != -2) return dp[c][x];
  if (c == l)
    if (x == 0)
      return dp[c][x] = 0;
    else
      return dp[c][x] = -1;
  int ans;
  if (a[c] == -1)
    ans = -1;
  else {
    if (x < (a[c] - c + 1 - l1))
      ans = -1;
    else {
      int h = rec(a[c] + 1, x - (a[c] - c + 1 - l1));
      if (h == -1)
        ans = -1;
      else
        ans = h + 1;
    }
  }
  ans = max(ans, rec(c + 1, x));
  if (x != 0) ans = max(ans, rec(c + 1, x - 1));
  return dp[c][x] = ans;
}
int main() {
  ios::sync_with_stdio(false);
  ifstream inf("input.txt");
  for (int i = 0; i <= 2001 - 1; i++)
    for (int i1 = 0; i1 <= 2001 - 1; i1++) dp[i][i1] = -2;
  cin >> s >> p;
  l = s.size(), l1 = p.size();
  for (int i = 0; i <= l - 1; i++) {
    int j = i, h = 0;
    while (j < l && h < l1) {
      if (s[j] == p[h]) h++;
      j++;
    }
    if (j == l && h < l1)
      a[i] = -1;
    else
      a[i] = j - 1;
  }
  for (int i = 0; i <= l; i++) {
    int h = rec(0, i);
    if (h == -1)
      cout << 0;
    else
      cout << h;
    cout << ' ';
  }
}
