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 e[2048], a[2048], f[2048], g[2048][2048];
void cm(int& x, int y) {
  if (x > y) x = y;
}
int main() {
  cin >> s >> p;
  for (size_t i = 0, j, k; i < s.size(); ++i) {
    for (j = i, k = 0; j < s.size() && k < p.size(); ++j)
      if (s[j] == p[k]) ++k;
    e[i] = k == p.size() ? j : 0;
  }
  for (int i = 0; i < s.size(); ++i) {
    f[i + 1] = max(f[i + 1], f[i]);
    if (e[i]) f[e[i]] = max(f[e[i]], (int)(f[i] + 65536 - e[i] + i + p.size()));
  }
  size_t x = f[s.size()] + 65535 >> 16, y = x * 65536 - f[s.size()];
  for (size_t i = y; i <= s.size(); ++i)
    a[i] = min(x * p.size(), s.size() - i) / p.size();
  for (size_t i = 1; i < x; ++i) {
    fill(g[i], g[i] + s.size() + 1, 999999);
    for (size_t j = 0, o = i - 1; j < s.size(); ++j) {
      if (e[j]) cm(g[i][e[j]], g[o][j] + e[j] - j - p.size());
      cm(g[i][j + 1], g[i][j]);
    }
    int t = g[i][s.size()];
    for (int j = t; j < y; ++j) a[j] = max(a[j], (int)i);
  }
  for (size_t i = 0; i <= s.size(); ++i)
    printf("%d%c", a[i], " \n"[i == s.size()]);
}
